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.

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

\[ \frac{dy}{dx}=v+x\frac{dv}{dx} \]

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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 \[ v=\frac{y}{x} \] \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ v=\ln|x|+C \] \[ y=x(\ln|x|+C) \]

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 \[ \int \frac{1}{y}dy=\int (1+x)dx \] \[ \ln |y|=x+\frac{x^2}{2}+C \] \[ y=Ce^{x+x^2/2} \]

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

\[ \frac{dy}{y}=e^x dx \] \[ \ln y=e^x+C \] \[ y=Ce^{e^x} \]

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 = LL^T

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

• Forward substitution: LY = B
• Backward substitution: L^TX = 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

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

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

This represents

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