Assignment: Probability and Statistics Basic

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Probability Problems with Detailed Solutions

Click each question to expand the detailed interpretation and solution.

Question 1: Two unbiased dice are thrown
Two fair dice are rolled together. Find the probability that:
(i) the sum is even
(ii) the sum is less than 5
(iii) the sum is at most 12
Since each die has 6 possible outcomes, the total number of outcomes is: $$6\times6=36$$ We calculate the number of favorable outcomes for each condition.
(i) Sum is even
The sum is even when:
  • both numbers are even
  • both numbers are odd
Number of even numbers on a die = 3
Number of odd numbers on a die = 3 $$3\times3+3\times3=18$$ Therefore, $$P(\text{sum is even})=\frac{18}{36}=\frac12$$
(ii) Sum is less than 5
Possible sums are 2, 3 and 4. Number of favorable outcomes: $$1+2+3=6$$ Hence, $$P(\text{sum}<5)=\frac{6}{36}=\frac16$$
(iii) Sum is at most 12
The largest possible sum of two dice is 12, so all outcomes are favorable. $$P(\text{sum}\le 12)=1$$
Final Answers:
\( P(\text{even})=\frac12 \)
\( P(<5)=\frac16 \)
\( P(\le12)=1 \)
Question 2: Four cards are drawn from a pack of 52 cards
Four cards are selected at random from a standard deck of 52 cards. Find the probability that:
(i) all four are aces
(ii) exactly three are kings and one is a queen
(iii) exactly three are black and one is red
The total number of ways of drawing 4 cards from 52 is: $$^{52}C_4=270725$$ Each required event is counted separately and divided by total possible selections.
(i) All four are aces
Number of ways: $$^{4}C_4=1$$ Therefore, $$P=\frac{1}{270725}$$
(ii) Exactly three kings and one queen
Ways to choose 3 kings: $$^{4}C_3=4$$ Ways to choose 1 queen: $$^{4}C_1=4$$ Total favorable ways: $$4\times4=16$$ Thus, $$P=\frac{16}{270725}$$
(iii) Exactly three black and one red
Black cards = 26, Red cards = 26 $$^{26}C_3\times^{26}C_1$$ Hence, $$P=\frac{^{26}C_3\times^{26}C_1}{^{52}C_4}=\frac{104}{417}$$
Final Answers:
\( \frac1{270725} \)
\( \frac{16}{270725} \)
\( \frac{104}{417} \)
Question 3: Exactly 8 cards of the same suit in a bridge hand
In a bridge hand of 13 cards, find the probability of getting exactly 8 cards from one suit.
A bridge hand consists of 13 cards drawn from 52 cards. We first count all possible bridge hands, then count the hands having exactly 8 cards from one suit.
Total number of bridge hands: $$^{52}C_{13}$$ Choose one suit from 4: $$^{4}C_1=4$$ Choose 8 cards from the 13 cards in that suit: $$^{13}C_8$$ Choose remaining 5 cards from remaining 39 cards: $$^{39}C_5$$ Thus, $$P=\frac{4\binom{13}{8}\binom{39}{5}}{\binom{52}{13}}$$
Final Answer:
\( \frac{4\binom{13}{8}\binom{39}{5}}{\binom{52}{13}} \)
Question 4: At least one additional heart
A man is dealt 3 hearts from a standard pack of 52 cards. He is then given 4 more cards. Find the probability that at least one of these additional 4 cards is also a heart.
Since the player already has 3 hearts, there are only 10 hearts left in the remaining 49 cards. Instead of directly finding the probability of getting at least one heart, it is easier to find the probability of getting no hearts in the next 4 cards, and subtract from 1.
Remaining cards = 49
Remaining hearts = 10
Remaining non-hearts = 39

Probability of no hearts in 4 additional cards: $$P(\text{no heart})=\frac{{^{39}C_4}}{{^{49}C_4}}$$ Therefore, $$P(\text{at least one heart})=1-\frac{{^{39}C_4}}{{^{49}C_4}}$$ $$P=1-\frac{82251}{211876}=0.6117$$
Final Answer:
\( P = 0.6117 \)
Question 5: Vowels together in "BALLOON"
If the letters of the word BALLOON are arranged at random, what is the probability that all the vowels come together?
The word BALLOON has 7 letters with repeated letters:
L appears twice, and O appears twice.
We first find the total number of arrangements of the word. Then we treat the vowels AOO as one block and count the favorable arrangements.
Total arrangements of BALLOON: $$\frac{7!}{2!2!}=1260$$ Vowels are A, O, O. Treat them as one block: $$(AOO), B, L, L, N$$ So we have 5 objects, with L repeated twice. Number of arrangements: $$\frac{5!}{2!}=60$$ The vowels AOO can be arranged among themselves in: $$\frac{3!}{2!}=3$$ Therefore favorable arrangements: $$60\times3=180$$ Hence, $$P=\frac{180}{1260}=\frac17$$
Final Answer:
\( P = \frac17 \)
Question 6: Word arrangement problems
(a) If the letters of the word EDUCATION are arranged at random, find the probability that there are exactly 3 letters between E and N.

(b) If the letters of the word PARALLEL are arranged at random, find the probability that all three L’s come together.
In both parts, we calculate:
Probability = Favorable arrangements / Total arrangements

For (a), we place E and N exactly 4 positions apart.
For (b), we treat the three L’s as one single block.
(a) Word EDUCATION

EDUCATION has 9 distinct letters. Total arrangements: $$9!$$ To have exactly 3 letters between E and N, the pair (E,N) must occupy: $$(1,5),(2,6),(3,7),(4,8),(5,9)$$ So number of position choices = 5 E and N can interchange in: $$2!$$ Remaining 7 letters can be arranged in: $$7!$$ Favorable arrangements: $$5\times2\times7!$$ Hence, $$P=\frac{5\times2\times7!}{9!}=\frac{5}{36}$$

(b) Word PARALLEL

Letters: P, A, R, A, L, L, E, L Total arrangements: $$\frac{8!}{2!3!}$$ Treat the three L’s as one block: $$(LLL), P, A, R, A, E$$ This gives 6 objects with A repeated twice. Favorable arrangements: $$\frac{6!}{2!}$$ Therefore, $$P=\frac{\frac{6!}{2!}}{\frac{8!}{2!3!}}=\frac{3}{28}$$
Final Answers:
(a) \( \frac{5}{36} \)
(b) \( \frac{3}{28} \)
Question 7: Particular pair of books on a shelf
Ten books are placed at random on a shelf. Find the probability that a particular pair of books shall be:

(i) Always together
(ii) Never together
We first count the total number of ways of arranging 10 books on a shelf: $$10!$$ For part (i), we treat the particular pair as one unit. For part (ii), we subtract the probability of being together from 1.
(i) Probability that the pair is always together

Treat the pair as one single block. Then the total objects are:
8 other books + 1 block = 9 objects Number of arrangements of these 9 objects: $$9!$$ The two books inside the pair can be arranged in: $$2!$$ Therefore favorable arrangements: $$9!\times2!$$ Probability: $$P=\frac{9!\times2!}{10!}=\frac{2}{10}=\frac15$$

(ii) Probability that the pair is never together

$$P(\text{never together})=1-\frac15=\frac45$$
Final Answers:
(i) \( \frac15 \)
(ii) \( \frac45 \)
Question 8: Four-digit number formed from 1,2,3,4,5
A four-digit number is formed using the digits 1, 2, 3, 4, 5 without repetition. Find the probability that the number is even.
A number is even if its last digit is even. Among the digits {1,2,3,4,5}, only 2 and 4 are even. We first find the total number of 4-digit numbers, then count how many end with an even digit.
Total 4-digit numbers formed from 5 digits: $$^5P_4=5\times4\times3\times2=120$$ To form an even number, the last digit must be 2 or 4. So there are 2 choices for the last digit. The remaining 3 places can be filled by the remaining 4 digits: $$^4P_3=4\times3\times2=24$$ Favorable numbers: $$2\times24=48$$ Therefore, $$P=\frac{48}{120}=\frac25$$
Final Answer:
\( \frac25 \)
Question 9: Comparing chances of throwing 6, 7 and 10
Compare the chances of:
(i) throwing a 6 with one die
(ii) throwing a 7 with two dice
(iii) throwing a 10 with three dice
We calculate the probability of each event:
- Getting 6 on one die
- Getting sum 7 on two dice
- Getting sum 10 on three dice
(i) Throwing a 6 with one die $$P=\frac16$$

(ii) Throwing a sum of 7 with two dice Total outcomes: $$6\times6=36$$ Favorable outcomes: $$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$$ Number of favorable outcomes = 6 $$P=\frac{6}{36}=\frac16$$

(iii) Throwing a sum of 10 with three dice Total outcomes: $$6^3=216$$ Number of favorable outcomes for sum 10 = 27 $$P=\frac{27}{216}=\frac18$$
Final Answers:
(i) \( \frac16 \)
(ii) \( \frac16 \)
(iii) \( \frac18 \)
Question 10: Product of two non-negative numbers
The sum of two non-negative numbers is \(S\). Find the probability that their product is at least half of the maximum possible product.
Let the numbers be \(x\) and \(S-x\). Their product is: $$x(S-x)$$ The maximum product occurs when the numbers are equal: $$x=\frac{S}{2}$$ Maximum product: $$\frac{S^2}{4}$$ We want the probability that: $$x(S-x)\ge\frac12\times\frac{S^2}{4}=\frac{S^2}{8}$$
Solve: $$x(S-x)\ge\frac{S^2}{8}$$ $$Sx-x^2\ge\frac{S^2}{8}$$ Rearranging: $$x^2-Sx+\frac{S^2}{8}\le0$$ Solving gives: $$x\in\left[\frac{S}{2}\left(1-\frac{1}{\sqrt2}\right),\frac{S}{2}\left(1+\frac{1}{\sqrt2}\right)\right]$$ Length of favorable interval: $$\frac{S}{\sqrt2}$$ Total interval length: $$S$$ Therefore, $$P=\frac{S/\sqrt2}{S}=\frac{1}{\sqrt2}$$
Final Answer:
\( \frac{1}{\sqrt2} \)
Question 11: Sum of two dice
If two dice are thrown, find the probability that the sum is:
(i) less than 5
(ii) a multiple of 3
When two dice are thrown, total possible outcomes: $$6\times6=36$$ We count favorable outcomes for each event.
(i) Sum less than 5 Possible sums: 2, 3, 4 Number of outcomes: $$1+2+3=6$$ Therefore, $$P=\frac{6}{36}=\frac16$$

(ii) Sum is a multiple of 3 Multiples of 3 are 3, 6, 9, 12 Number of outcomes: $$2+5+4+1=12$$ Therefore, $$P=\frac{12}{36}=\frac13$$
Final Answers:
(i) \( \frac16 \)
(ii) \( \frac13 \)
Question 12: Heart or Face Card
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is either a Heart or a Face card (King, Queen, Jack).
We use: $$P(H\cup F)=P(H)+P(F)-P(H\cap F)$$ Hearts = 13 cards, Face cards = 12 cards, Heart face cards = 3 cards.
Favorable cards: $$13+12-3=22$$ Therefore: $$P=\frac{22}{52}=\frac{11}{26}$$
Final Answer:
\( \frac{11}{26} \)
Question 13: Marbles in a bag
A bag contains 5 red and 3 blue marbles. If 3 marbles are drawn at random, find the probability that at least one is blue.
Instead of directly finding "at least one blue", we use complement: $$P(\text{at least one blue})=1-P(\text{all red})$$
Total ways: $$^8C_3$$ Ways to draw all red: $$^5C_3$$ Therefore: $$P=1-\frac{^5C_3}{^8C_3}$$ $$P=1-\frac{10}{56}=\frac{23}{28}$$
Final Answer:
\( \frac{23}{28} \)
Question 14: Marksmen hitting a target
Three marksmen take one shot each at a target. Their probabilities of hitting are 0.6, 0.5, and 0.8. Find the probability that the target is hit at least once.
It is easier to find the probability that all miss, then subtract from 1.
Probability all miss: $$(1-0.6)(1-0.5)(1-0.8)$$ $$=0.4\times0.5\times0.2=0.04$$ Therefore: $$P(\text{at least one hit})=1-0.04=0.96$$
Final Answer:
\( 0.96 \)
Question 15: Showing \(P(A\cap B)\ge 0.5\)
If \(P(A)=0.7\) and \(P(B)=0.8\), show that: $$P(A\cap B)\ge0.5$$
Use: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ Since \(P(A\cup B)\le1\), we get a lower bound for \(P(A\cap B)\).
$$0.7+0.8-P(A\cap B)\le1$$ $$1.5-P(A\cap B)\le1$$ $$0.5\le P(A\cap B)$$ Hence: $$P(A\cap B)\ge0.5$$
Final Answer:
\( P(A\cap B)\ge0.5 \)
Question 16: Last fused bulb
A box contains 10 bulbs, 3 fused. Bulbs are tested one by one without replacement. Find the probability that the 6th bulb tested is the last fused one.
For the 6th bulb to be the last fused: - exactly 2 fused among first 5 bulbs - 6th bulb is fused
Probability: $$\frac{^3C_2\cdot ^7C_3}{^10C_6}$$ $$=\frac{3\times35}{210}=\frac12$$
Final Answer:
\( \frac12 \)
Question 17: First death among n men
\(p\) is the probability that a man aged \(x\) dies in a year. Find the probability that among \(n\) men, \(A_1\) dies in a year and is the first to die.
If exactly \(r\) men die, then: - \(A_1\) must be one of them - \(A_1\) must be the first among the \(r\) deaths
The required probability is: $$\frac1n\left[1-(1-p)^n\right]$$
Final Answer:
\( \frac1n\left[1-(1-p)^n\right] \)
Question 18: Contradicting statements
Odds that A speaks truth are 4:1 and odds that B speaks truth are 3:2. Find the percentage of cases in which they contradict each other.
They contradict when: - A tells truth, B lies - A lies, B tells truth
$$P(A_T)=\frac45,\quad P(B_T)=\frac35$$ Contradiction: $$\frac45\cdot\frac25+\frac15\cdot\frac35$$ $$=\frac{8}{25}+\frac{3}{25}=\frac{11}{25}=44\%$$
Final Answer:
\(44\%\)
Question 19: Sustainable building design
Architects A, B, and C are selected in ratio 3:2:5. Their probabilities of sustainable design are 0.4, 0.6, and 0.9. Find the probability that the final project is sustainable.
Use weighted probability: $$P(S)=P(A)P(S|A)+P(B)P(S|B)+P(C)P(S|C)$$
$$P(A)=\frac3{10},\quad P(B)=\frac2{10},\quad P(C)=\frac5{10}$$ $$P(S)=\frac3{10}(0.4)+\frac2{10}(0.6)+\frac5{10}(0.9)$$ $$=0.12+0.12+0.45=0.69$$
Final Answer:
\(0.69\)
Question 20: Screen defect probability
Lines X, Y, Z produce 50%, 30%, 20% of smartphones. Their defect rates are 1%, 2%, 3%. If a phone has a screen defect, find the probability it was manufactured by line X.
Use Bayes' theorem: $$P(X|D)=\frac{P(X)P(D|X)}{P(D)}$$
$$P(D)=0.5(0.01)+0.3(0.02)+0.2(0.03)=0.017$$ $$P(X|D)=\frac{0.5\times0.01}{0.017}=\frac{5}{17}$$
Final Answer:
\( \frac{5}{17} \)
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Some Questions on Differential Equations

Differential Equations: Detailed Step-by-Step Solutions

In this article, we solve three important differential equations step by step using standard methods such as complementary function and particular integral, integrating factor, and separation of variables.

1) Solve \(y''+4y=\sin(3x)\)

This is a linear differential equation with constant coefficients:

\[ y''+4y=\sin(3x) \]

The general solution is:

\[ y=y_c+y_p \]

where \(y_c\) is the complementary function and \(y_p\) is the particular integral.

Step 1: Complementary Function

The auxiliary equation is:

\[ m^2+4=0 \] \[ m^2=-4 \] \[ m=\pm 2i \]

Therefore, the complementary function is:

\[ y_c=C_1\cos 2x+C_2\sin 2x \]

Step 2: Particular Integral

\[ (D^2+4)y=\sin 3x \] \[ y_p=\frac{1}{D^2+4}\sin 3x \] Using: \[ f(D)\sin ax=f(-a^2)\sin ax \] \[ y_p=\frac{1}{-9+4}\sin 3x \] \[ y_p=-\frac15\sin 3x \]

Step 3: General Solution

\[ y=C_1\cos 2x+C_2\sin 2x-\frac15\sin 3x \]
Final Answer: \[ \boxed{y=C_1\cos 2x+C_2\sin 2x-\frac15\sin 3x} \]
2) Solve \(y'+y\tan x=\sin 2x\)

This is a first-order linear differential equation:

\[ \frac{dy}{dx}+Py=Q \] where \[ P=\tan x,\quad Q=\sin 2x \]

Step 1: Integrating Factor

\[ IF=e^{\int Pdx} \] \[ IF=e^{\int \tan x\,dx} \] \[ IF=e^{\log\sec x} \] \[ IF=\sec x \]

Step 2: Multiply throughout by IF

\[ \sec x\frac{dy}{dx}+y\sec x\tan x=\sin 2x\sec x \] The left-hand side becomes: \[ \frac{d}{dx}(y\sec x) \] Thus, \[ \frac{d}{dx}(y\sec x)=\sin 2x\sec x \] Since \[ \sin 2x=2\sin x\cos x \] we get \[ \sin 2x\sec x=2\sin x \] Hence, \[ \frac{d}{dx}(y\sec x)=2\sin x \]

Step 3: Integrate

\[ y\sec x=\int 2\sin x\,dx \] \[ y\sec x=-2\cos x+C \] Multiplying by \(\cos x\), \[ y=-2\cos^2x+C\cos x \]
Final Answer: \[ \boxed{y=-2\cos^2x+C\cos x} \]
3) Solve \(yy'+25x=0\)

Given:

\[ y\frac{dy}{dx}+25x=0 \] Rearranging, \[ y\frac{dy}{dx}=-25x \] Separate variables: \[ y\,dy=-25x\,dx \]

Step 1: Integrate both sides

\[ \int y\,dy=\int -25x\,dx \] \[ \frac{y^2}{2}=-\frac{25x^2}{2}+C \] Multiplying by 2: \[ y^2=-25x^2+C \] or, \[ y^2+25x^2=C \]
Final Answer: \[ \boxed{y^2+25x^2=C} \]

Conclusion

These examples demonstrate three important methods for solving differential equations: complementary function & particular integral, integrating factor, and separation of variables. Mastering these techniques helps in solving a wide variety of ordinary differential equations.

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Chapter 3: First Order Homogeneous Differential Equations

First Order Homogeneous Differential Equations (Solved Examples)

A differential equation of the form

\[ \frac{dy}{dx}=f\left(\frac{y}{x}\right) \]

is called a homogeneous differential equation. We use the substitution

\[ y=vx \quad \text{or} \quad v=\frac{y}{x} \]

so that

\[ \frac{dy}{dx}=v+x\frac{dv}{dx} \]
Example 1: Solve \( \frac{dy}{dx}=\frac{x+y}{x} \)
View Solution
\[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ \frac{dv}{dx}=\frac{1}{x} \] Integrate \[ v=\ln|x|+C \] Since \(v=\frac{y}{x}\) \[ \frac{y}{x}=\ln|x|+C \] \[ y=x(\ln|x|+C) \]
Example 2: Solve \( \frac{dy}{dx}=\frac{x+y}{x-y} \)
View Solution
Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=\frac{1+v}{1-v} \] \[ x\frac{dv}{dx}=\frac{1+v}{1-v}-v \] \[ x\frac{dv}{dx}=\frac{1+v^2}{1-v} \] Separate variables \[ \frac{1-v}{1+v^2}dv=\frac{dx}{x} \] Integrate \[ \tan^{-1}v-\frac12\ln(1+v^2)=\ln|x|+C \]
Example 3: Solve \( \frac{dy}{dx}=\frac{x-y}{x+y} \)
View Solution
Let \[ v=\frac{y}{x} \] \[ y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=\frac{1-v}{1+v} \] \[ x\frac{dv}{dx}=\frac{1-v}{1+v}-v \] Solve by separation of variables.
Example 4: Solve \( \frac{dy}{dx}=\frac{y}{x}+1 \)
View Solution
\[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ v=\ln|x|+C \] \[ y=x(\ln|x|+C) \]
Example 5: Solve \( \frac{dy}{dx}=\frac{x^2+y^2}{xy} \)
View Solution
\[ \frac{dy}{dx}=\frac{x}{y}+\frac{y}{x} \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=\frac{1}{v}+v \] \[ x\frac{dv}{dx}=\frac{1}{v} \] \[ v\,dv=\frac{dx}{x} \] Integrate \[ \frac{v^2}{2}=\ln|x|+C \] \[ \left(\frac{y}{x}\right)^2=2\ln|x|+C \]
Example 6: Solve \( \frac{dy}{dx}=\frac{x+y}{y} \)
View Solution
\[ \frac{dy}{dx}=\frac{x}{y}+1 \] Let \[ v=\frac{y}{x} \] Substitute \(y=vx\) and solve the resulting separable equation.
Example 7: Solve \( \frac{dy}{dx}=\frac{y-x}{x} \)
View Solution
\[ \frac{dy}{dx}=\frac{y}{x}-1 \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=v-1 \] \[ x\frac{dv}{dx}=-1 \] \[ v=-\ln|x|+C \] \[ \frac{y}{x}=-\ln|x|+C \]
Example 8: Solve \( \frac{dy}{dx}=\frac{x^2-y^2}{xy} \)
View Solution
\[ \frac{dy}{dx}=\frac{x}{y}-\frac{y}{x} \] Let \[ y=vx \] Substitute and solve the separable equation.
Example 9: Solve \( \frac{dy}{dx}=\left(\frac{y}{x}\right)^2 \)
View Solution
Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=v^2 \] \[ x\frac{dv}{dx}=v^2-v \] Separate variables \[ \frac{dv}{v(v-1)}=\frac{dx}{x} \] Integrate using partial fractions.
Example 10: Solve \( \frac{dy}{dx}=\frac{x^2+xy}{x^2} \)
View Solution
\[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ v=\ln|x|+C \] \[ y=x(\ln|x|+C) \]
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ODE Chapter 1: First Order Differential Equations using Separation of Variables

First Order Differential Equations using Separation of Variables

Below are solved examples of first order differential equations using the separation of variables method.

Example 1: Solve \( \frac{dy}{dx}=3x^2 \)
View Solution
Separate variables \[ dy = 3x^2 dx \] Integrate both sides \[ \int dy = \int 3x^2 dx \] \[ y = x^3 + C \]
Example 2: Solve \( \frac{dy}{dx}=xy \)
View Solution
Separate variables \[ \frac{dy}{y}=x\,dx \] Integrate \[ \int \frac{1}{y}dy = \int x\,dx \] \[ \ln |y| = \frac{x^2}{2}+C \] \[ y = Ce^{x^2/2} \]
Example 3: Solve \( \frac{dy}{dx}=\frac{x}{y} \)
View Solution
Separate variables \[ y\,dy = x\,dx \] Integrate \[ \int y\,dy = \int x\,dx \] \[ \frac{y^2}{2} = \frac{x^2}{2} + C \] \[ y^2 = x^2 + C \]
Example 4: Solve \( \frac{dy}{dx}=y^2 \)
View Solution
Separate variables \[ \frac{dy}{y^2}=dx \] Integrate \[ \int y^{-2}dy = \int dx \] \[ -\frac{1}{y}=x+C \] \[ y=\frac{1}{C-x} \]
Example 5: Solve \( \frac{dy}{dx}=\frac{1+y^2}{x} \)
View Solution
Separate variables \[ \frac{dy}{1+y^2}=\frac{dx}{x} \] Integrate \[ \int \frac{dy}{1+y^2}=\int \frac{dx}{x} \] \[ \tan^{-1}y=\ln |x|+C \] \[ y=\tan(\ln |x|+C) \]
Example 6: Solve \( \frac{dy}{dx}=y\cos x \)
View Solution
Separate variables \[ \frac{dy}{y}=\cos x\,dx \] Integrate \[ \int \frac{1}{y}dy=\int \cos x\,dx \] \[ \ln |y|=\sin x + C \] \[ y=Ce^{\sin x} \]
Example 7: Solve \( \frac{dy}{dx}=\frac{y}{1+x} \)
View Solution
Separate variables \[ \frac{dy}{y}=\frac{dx}{1+x} \] Integrate \[ \int \frac{1}{y}dy=\int \frac{1}{1+x}dx \] \[ \ln |y|=\ln |1+x|+C \] \[ y=C(1+x) \]
Example 8: Solve \( \frac{dy}{dx}=xe^{y} \)
View Solution
Separate variables \[ e^{-y}dy=x\,dx \] Integrate \[ \int e^{-y}dy=\int x\,dx \] \[ -e^{-y}=\frac{x^2}{2}+C \]
Example 9: Solve \( \frac{dy}{dx}=\frac{x^2}{y+1} \)
View Solution
Separate variables \[ (y+1)dy=x^2dx \] Integrate \[ \int (y+1)dy=\int x^2dx \] \[ \frac{y^2}{2}+y=\frac{x^3}{3}+C \]
Example 10: Solve \( \frac{dy}{dx}=y(1+x) \)
View Solution
Separate variables \[ \frac{dy}{y}=(1+x)dx \] Integrate \[ \int \frac{1}{y}dy=\int (1+x)dx \] \[ \ln |y|=x+\frac{x^2}{2}+C \] \[ y=Ce^{x+x^2/2} \]
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Differential Equation: Application

First Order Differential Equations using Separation of Variables

Advanced Application Problems (Separation of Variables)

The following examples illustrate applications of first order differential equations in population growth, cooling law, chemical reactions, and physics.

Example 11 (Population Growth): If population grows according to \( \frac{dP}{dt}=kP \), find \(P(t)\).
View Solution
Separate variables \[ \frac{dP}{P}=k\,dt \] Integrate \[ \int \frac{dP}{P}=\int k\,dt \] \[ \ln P = kt + C \] \[ P = Ce^{kt} \]
Example 12 (Newton's Law of Cooling): \[ \frac{dT}{dt}=-k(T-T_s) \] Find \(T(t)\).
View Solution
Separate variables \[ \frac{dT}{T-T_s}=-k\,dt \] Integrate \[ \ln |T-T_s|=-kt+C \] \[ T-T_s=Ce^{-kt} \] \[ T=T_s+Ce^{-kt} \]
Example 13 (Radioactive Decay): Solve \( \frac{dN}{dt}=-kN \)
View Solution
\[ \frac{dN}{N}=-k\,dt \] Integrate \[ \ln N=-kt+C \] \[ N=Ce^{-kt} \]
Example 14 (Chemical Reaction): Solve \[ \frac{dy}{dx}=ky(1-y) \]
View Solution
Separate variables \[ \frac{dy}{y(1-y)}=k\,dx \] Using partial fractions \[ \frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y} \] Integrate \[ \ln|y|-\ln|1-y|=kx+C \]
Example 15 (Logistic Population Model) \[ \frac{dP}{dt}=rP\left(1-\frac{P}{K}\right) \]
View Solution
Separate variables \[ \frac{dP}{P(1-P/K)}=r\,dt \] Integrate using partial fractions \[ \ln\frac{P}{K-P}=rt+C \]
Example 16: Solve \[ \frac{dy}{dx}=x\sqrt{y} \]
View Solution
Separate variables \[ \frac{dy}{\sqrt{y}}=x\,dx \] Integrate \[ 2\sqrt{y}=\frac{x^2}{2}+C \]
Example 17 (Bacterial Growth): \[ \frac{dB}{dt}=0.3B \]
View Solution
\[ \frac{dB}{B}=0.3dt \] Integrate \[ \ln B=0.3t+C \] \[ B=Ce^{0.3t} \]
Example 18: Solve \[ \frac{dy}{dx}=\frac{y}{x^2} \]
View Solution
\[ \frac{dy}{y}=\frac{dx}{x^2} \] Integrate \[ \ln y=-\frac{1}{x}+C \]
Example 19 (Evaporation Model): \[ \frac{dV}{dt}=-kV \]
View Solution
\[ \frac{dV}{V}=-kdt \] Integrate \[ \ln V=-kt+C \] \[ V=Ce^{-kt} \]
Example 20: Solve \[ \frac{dy}{dx}=\frac{x}{1+y} \]
View Solution
Separate variables \[ (1+y)dy=x\,dx \] Integrate \[ y+\frac{y^2}{2}=\frac{x^2}{2}+C \]
Example 21 (Compound Interest Model) \[ \frac{dA}{dt}=rA \]
View Solution
\[ \frac{dA}{A}=rdt \] \[ \ln A=rt+C \] \[ A=Ce^{rt} \]
Example 22: Solve \[ \frac{dy}{dx}=y^3 \]
View Solution
\[ \frac{dy}{y^3}=dx \] Integrate \[ -\frac{1}{2y^2}=x+C \]
Example 23 (Drug Elimination Model) \[ \frac{dD}{dt}=-kD \]
View Solution
\[ \frac{dD}{D}=-kdt \] \[ D=Ce^{-kt} \]
Example 24: Solve \[ \frac{dy}{dx}=\sin x \, y \]
View Solution
\[ \frac{dy}{y}=\sin x dx \] \[ \ln y=-\cos x+C \] \[ y=Ce^{-\cos x} \]
Example 25: Solve \[ \frac{dy}{dx}=y\tan x \]
View Solution
\[ \frac{dy}{y}=\tan x dx \] \[ \ln y=-\ln|\cos x|+C \] \[ y=C\sec x \]
Example 26: Solve \[ \frac{dy}{dx}=x^2y \]
View Solution
\[ \frac{dy}{y}=x^2dx \] \[ \ln y=\frac{x^3}{3}+C \]
Example 27: Solve \[ \frac{dy}{dx}=\frac{y}{\sqrt{x}} \]
View Solution
\[ \frac{dy}{y}=\frac{dx}{\sqrt{x}} \] \[ \ln y=2\sqrt{x}+C \]
Example 28 (Cooling Problem): \[ \frac{dT}{dt}=-k(T-20) \]
View Solution
\[ \frac{dT}{T-20}=-kdt \] \[ \ln|T-20|=-kt+C \]
Example 29: Solve \[ \frac{dy}{dx}=\frac{x^2}{y^2} \]
View Solution
\[ y^2dy=x^2dx \] \[ \frac{y^3}{3}=\frac{x^3}{3}+C \]
Example 30: Solve \[ \frac{dy}{dx}=e^x y \]
View Solution
\[ \frac{dy}{y}=e^x dx \] \[ \ln y=e^x+C \] \[ y=Ce^{e^x} \]
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Cholesky Decomposition Method

Cholesky Decomposition Method – Solved Problems

The Cholesky Decomposition Method is used to solve systems of linear equations when the coefficient matrix is symmetric and positive definite. The matrix is decomposed as

A = LLT

where L is a lower triangular matrix. The solution is obtained in two steps:

  • Forward substitution: LY = B
  • Backward substitution: LTX = Y
Problem 1

Solve using Cholesky decomposition

4x + 2y = 6
2x + 3y = 7

Matrix form

A = [4 2 2 3]

L = [2 0 1 √2]

Forward substitution:

2y₁ = 6 y₁ = 3
3 + √2 y₂ = 7
y₂ = 2√2

Backward substitution:

√2 y = 2√2 y = 2
2x + 2 = 3 x = 1/2

Solution: x = 1/2 y = 2
Problem 2

Solve
4x + 2y + 2z = 8
2x + 5y + z = 3
2x + y + 3z = 5

Matrix
A = [4 2 2 2 5 1 2 1 3]

Cholesky factor
L = [2 0 0 1 2 0 1 0 √2]

Forward substitution gives
Y = [4 -1/2 1/√2]

Backward substitution
x = 2 y = -1 z = 1
Problem 3

Solve
9x + 3y = 12
3x + 5y = 7

Cholesky factor
L = [3 0 1 2]

Forward substitution
3y₁ = 12 y₁ = 4
4 + 2y₂ = 7
y₂ = 3/2

Backward substitution
x = 1 y = 1
Problem 4

Solve
16x + 4y = 20
4x + 10y = 18

L = [4 0 1 3]

Forward substitution
y₁ = 5
5 + 3y₂ = 18
y₂ = 3

Backward substitution gives
x = 1 y = 2
Problem 5

Solve
25x + 5y = 30
5x + 6y = 11

L = [5 0 1 √5]

Solution
x = 1 y = 1
Problem 6

Solve
4x + 2y + 2z = 2
2x + 10y + 4z = 6
2x + 4y + 9z = 5

After Cholesky decomposition and substitution:

x = 0 y = 1/2 z = 1/2
Problem 7

Solve
6x + 3y = 9
3x + 2y = 5

Using Cholesky decomposition

x = 1 y = 1
Problem 8

Solve
9x + 3y + 3z = 12
3x + 5y + z = 8
3x + y + 4z = 7

Solution

x = 1 y = 1 z = 1
Problem 9

Solve
16x + 8y = 24
8x + 5y = 13

Solution

x = 1 y = 1
Problem 10

Solve
4x + 2y + 2z = 6
2x + 5y + z = 7
2x + y + 3z = 5

Solution

x = 1 y = 1 z = 1
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Total Derivatives

Total Derivatives – Theory and Solved Problems
Definition
If \(z=f(x,y)\) where \(x\) and \(y\) depend on parameter \(t\), then the total derivative is
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \]
This formula is known as the **Chain Rule for multivariable functions**.

Visual Understanding of the Chain Rule

In problems involving total derivatives, variables often depend on other variables. The Chain Rule shows how a change in one variable affects another through intermediate variables. The following diagrams illustrate this dependency structure.

Diagram 1: Basic Chain Rule Structure

z x y t

This diagram represents:

\[ z = f(x,y), \quad x = x(t), \quad y = y(t) \] \[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \]

Diagram 2: Two-Level Dependency

z x y u v w

This represents

\[ z = f(x,y), \quad x = x(u,v), \quad y = y(w) \] \[ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} \] \[ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} \] \[ \frac{\partial z}{\partial w} = \frac{\partial z}{\partial y}\frac{\partial y}{\partial w} \]

Diagram 3: Polar Coordinate Chain Rule

z x y r θ

This corresponds to

\[ x = r\cos\theta, \qquad y = r\sin\theta \] \[ \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\cos\theta + \frac{\partial z}{\partial y}\sin\theta \]

Advanced Examples on Total Derivatives

Solved Examples
Example 1 Find \(dz/dt\) if
\(z = x^2y + y^3\)
where
\(x=t^2,\quad y=t^3\)
\[ \frac{\partial z}{\partial x}=2xy \] \[ \frac{\partial z}{\partial y}=x^2+3y^2 \] \[ \frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=3t^2 \] Using total derivative formula \[ \frac{dz}{dt}=(2xy)(2t)+(x^2+3y^2)(3t^2) \] Substitute \(x=t^2\), \(y=t^3\) \[ \frac{dz}{dt}=7t^6+9t^8 \]
Example 2 If
\(z=\ln(x^2+y^2)\)
where
\(x=r\cos\theta,\quad y=r\sin\theta\)
Find \(\partial z/\partial r\).

Since \[ x^2+y^2=r^2 \] Thus \[ z=\ln(r^2) \] Differentiate \[ \frac{\partial z}{\partial r}=\frac{2}{r} \]
Example 3 Find \(dz/dt\) if
\(z=e^{xy}\)
where
\(x=t^2,\quad y=t\)
\[ \frac{\partial z}{\partial x}=ye^{xy} \] \[ \frac{\partial z}{\partial y}=xe^{xy} \] \[ \frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=1 \] \[ \frac{dz}{dt}=ye^{xy}(2t)+xe^{xy} \] Substitute \(x=t^2\), \(y=t\).
Example 4. If \( z = x^2y + y^3 \), where \( x = t^2 \) and \( y = t^3 \), find \( \frac{dz}{dt} \).

\[ z = x^2y + y^3 \] Partial derivatives: \[ \frac{\partial z}{\partial x}=2xy \] \[ \frac{\partial z}{\partial y}=x^2+3y^2 \] Since \[ x=t^2, \quad y=t^3 \] \[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2 \] Total derivative: \[ \frac{dz}{dt} =\frac{\partial z}{\partial x}\frac{dx}{dt} +\frac{\partial z}{\partial y}\frac{dy}{dt} \] Substitute: \[ =2xy(2t)+(x^2+3y^2)(3t^2) \] Replacing \(x=t^2, y=t^3\): \[ =4t(t^2t^3)+3t^2(t^4+3t^6) \] \[ =4t^6+3t^6+9t^8 \] \[ \boxed{\frac{dz}{dt}=7t^6+9t^8} \]
Example 5. If \( z = e^{xy} \), where \( x = r\cos\theta \) and \( y = r\sin\theta \), find \( \frac{\partial z}{\partial r} \).

\[ z=e^{xy} \] Partial derivatives: \[ \frac{\partial z}{\partial x}=ye^{xy} \] \[ \frac{\partial z}{\partial y}=xe^{xy} \] Now \[ \frac{\partial x}{\partial r}=\cos\theta \] \[ \frac{\partial y}{\partial r}=\sin\theta \] Total derivative: \[ \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} \] Substitute: \[ =ye^{xy}\cos\theta + xe^{xy}\sin\theta \] Factor: \[ =e^{xy}(y\cos\theta+x\sin\theta) \] Replace \(x=r\cos\theta\), \(y=r\sin\theta\) \[ =e^{r^2\sin\theta\cos\theta}(r\sin\theta\cos\theta+r\cos\theta\sin\theta) \] \[ \boxed{\frac{\partial z}{\partial r}=2r\sin\theta\cos\theta \; e^{r^2\sin\theta\cos\theta}} \]
Example 6. If \( z = \ln(x^2+y^2) \), where \( x=t^2+1 \) and \( y=2t \), find \( \frac{dz}{dt} \).

\[ z=\ln(x^2+y^2) \] Partial derivatives: \[ \frac{\partial z}{\partial x}=\frac{2x}{x^2+y^2} \] \[ \frac{\partial z}{\partial y}=\frac{2y}{x^2+y^2} \] Now \[ \frac{dx}{dt}=2t \] \[ \frac{dy}{dt}=2 \] Total derivative: \[ \frac{dz}{dt} = \frac{2x}{x^2+y^2}(2t) + \frac{2y}{x^2+y^2}(2) \] \[ = \frac{4xt+4y}{x^2+y^2} \] Substitute \(x=t^2+1, y=2t\): \[ \boxed{\frac{dz}{dt}= \frac{4t(t^2+1)+8t}{(t^2+1)^2+4t^2}} \]
Example 7. If \( z = x^3y^2 \), where \( x = u+v \) and \( y = uv \), find \( \frac{\partial z}{\partial u} \).

\[ z=x^3y^2 \] Partial derivatives: \[ \frac{\partial z}{\partial x}=3x^2y^2 \] \[ \frac{\partial z}{\partial y}=2x^3y \] Now \[ \frac{\partial x}{\partial u}=1 \] \[ \frac{\partial y}{\partial u}=v \] Total derivative: \[ \frac{\partial z}{\partial u} = 3x^2y^2(1) + 2x^3y(v) \] \[ =3x^2y^2+2vx^3y \] Substitute \(x=u+v\), \(y=uv\): \[ \boxed{ \frac{\partial z}{\partial u} = 3(u+v)^2(u^2v^2)+2v(u+v)^3(uv) } \]
Example 8. If \( z = \sin(xy) \), where \( x=r^2 \), \( y=r^3 \), find \( \frac{dz}{dr} \).

\[ z=\sin(xy) \] \[ \frac{\partial z}{\partial x}=y\cos(xy) \] \[ \frac{\partial z}{\partial y}=x\cos(xy) \] Now \[ \frac{dx}{dr}=2r \] \[ \frac{dy}{dr}=3r^2 \] Total derivative: \[ \frac{dz}{dr} = y\cos(xy)(2r)+x\cos(xy)(3r^2) \] \[ =\cos(xy)(2ry+3r^2x) \] Substitute \(x=r^2, y=r^3\): \[ \boxed{\frac{dz}{dr} = \cos(r^5)(2r^4+3r^4) = 5r^4\cos(r^5)} \]
Example 9. If \( z = x^2+y^2 \), where \( x = \cos t \) and \( y = \sin t \), find \( \frac{dz}{dt} \).

\[ z=x^2+y^2 \] \[ \frac{\partial z}{\partial x}=2x \] \[ \frac{\partial z}{\partial y}=2y \] Now \[ \frac{dx}{dt}=-\sin t \] \[ \frac{dy}{dt}=\cos t \] Total derivative: \[ \frac{dz}{dt} = 2x(-\sin t)+2y(\cos t) \] Substitute \(x=\cos t, y=\sin t\): \[ =2\cos t(-\sin t)+2\sin t(\cos t) \] \[ \boxed{\frac{dz}{dt}=0} \]
Example 10. If \( z = e^{x+y} \), where \( x=t^2 \) and \( y=\ln t \), find \( \frac{dz}{dt} \).

\[ z=e^{x+y} \] Partial derivatives: \[ \frac{\partial z}{\partial x}=e^{x+y} \] \[ \frac{\partial z}{\partial y}=e^{x+y} \] Now \[ \frac{dx}{dt}=2t \] \[ \frac{dy}{dt}=\frac{1}{t} \] Total derivative: \[ \frac{dz}{dt} = e^{x+y}(2t)+e^{x+y}\left(\frac{1}{t}\right) \] \[ =e^{x+y}\left(2t+\frac{1}{t}\right) \] Substitute \(x=t^2\), \(y=\ln t\): \[ \boxed{\frac{dz}{dt}=e^{t^2+\ln t}\left(2t+\frac{1}{t}\right)} \]
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Assignment: Probability and Statistics Basic

Sticky Ad Probability Problems with Detailed Solutions Click each question to expand the detailed interpretation and solution. ...