Question

If the solution of the differential equation $x(x^2+1)\left(\frac{dy}{dx}\right)=y(1-x^2)+x^3\ln x$ is $\frac{y(x^2+1)}{x}=A{x}^2\ln x+B{x}^2+C,$ then which of the following is/are TRUE $?$
(where $C$ is an arbitrary constant)

• A. $A=\frac{1}{2}$
• B. $A=2B$
• C. $A=-2B$
• D. $B=\frac{1}{2}$

Answer 1

Answer: (A), (C)

$A=-2B$

The given equation can be rewitten as
$\frac{dy}{dx}+\frac{x^2-1}{x(x^2+1)}y=\frac{x^2\ln x}{(x^2+1)}\cdots \left(1\right)$ which is linear and $x>0$.
Also $P=\frac{x^2-1}{x(x^2+1)}$ and $Q=\frac{x^2\ln x}{(x^2+1)}$
Now,
$\underset{}{\overset{}{\int }}Pdx=\underset{}{\overset{}{\int }}[\frac{2x}{x^2+1}-\frac{1}{x}]dx$
$\left[\text{Resolving into partial fractions}\right]$
$=\ln (x^2+1)-\ln x$
$\therefore \text{I.F.}=e^{\ln \left|\frac{(x^2+1)}{x}\right|}=\frac{x^2+1}{x}$

Hence the required solution of equation $\left(1\right)$ is
$\frac{y(x^2+1)}{x}=\underset{}{\overset{}{\int }}\frac{(x^2+1)}{x}\frac{x^2\ln x}{(x^2+1)}dx+C$
$\Rightarrow \frac{y(x^2+1)}{x}=\underset{}{\overset{}{\int }}x\ln xdx+C$
On integrating By parts, we get:
$\Rightarrow \frac{y(x^2+1)}{x}=\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2+C$
$\Rightarrow A=\frac{1}{2};B=-\frac{1}{4}$
$\Rightarrow A=-2B$