Question

The solution $\left(1+x^2\right)\frac{dy}{dx}+2xy-4x^2=0$ is

• A. $3x(1+y^2)=4y^3+c$
• B. $3y(1+x^2)=4x^3+c$
• C. $3x(1-y^2)=4y^3+c$
• D. $3y(1+y^2)=4x^3+c$

Answer 1

The correct option is B

$3y(1+x^2)=4x^3+c$

Explanation for the correct answer:

Compute the required solution:

$\left(1+x^2\right)\frac{dy}{dx}+2xy-4x^2=0$

$\Rightarrow$ $\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{4x^2}{1+x^2}$ $...\left(i\right)$

Equation $\left(i\right)$ is a linear differential equation of the form

$\frac{dy}{dx}+Py=Q$ where, $P=\frac{2x}{1+x^2}$, $Q=\frac{4x^2}{1+x^2}$

Integrating factor $I.F=e^{\int Pdx}$

$\Rightarrow$ $I.F=e^{\int \frac{2x}{1+x^2}dx}$

Let $1+x^2=t$

$\therefore$ $2xdx=dt$

$\Rightarrow$$I.F=e^{\int \frac{dt}{t}}$

$\Rightarrow$$I.F=e^{\ln \left(t\right)}$

$\Rightarrow$$I.F=t=1+x^2$

The solution of the given differential equation is given as

$y\left(I.F\right)=\int \left(I.F\right)Qdx+c_1$

$\Rightarrow$ $y\left(1+x^2\right)=\int \left(1+x^2\right)\frac{4x^2}{1+x^2}dx+c_1$

$\Rightarrow$ $y\left(1+x^2\right)=\int 4x^{2}dx+c_1$

$\Rightarrow$ $y\left(1+x^2\right)=4\left[\frac{x^3}{3}\right]+c_1$

$\Rightarrow$$3y\left(1+x^2\right)=4x^3+c$

Hence the solution of the given differential equation is $3y\left(1+x^2\right)=4x^3+c$.

Hence, option $\left(B\right)$ is the correct answer.