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

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