Question

Solve the differential equations:
$(1+y+x^{2}y)dx=(x+x^3)dy$

• A. $y=cx-x\arctan x-1$
• B. $y=cx-\arctan x-1$
• C. $xy=c-arccotx$
• D. $xy=c+arccotx$

Answer 1

The correct option is A $y=cx-x\arctan x-1$

$(1+y+x^{2}y)dx=(x+x^3)dy$
Given differential eqn can be written as
$\frac{dy}{dx}=\frac{1+y+x^{2}y}{(x+x^3)}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{x+x^3}+\frac{y(1+x^2)}{x+x^3}$
$\Rightarrow \frac{dy}{dx}-\frac{y}{x}=\frac{1}{x+x^3}$
which is a linear differential equation.
Here, $P=-\frac{1}{x}$, $Q=\frac{1}{x+x^3}$
Integrating Factor $I.F.=e^{\int Pdx}$
$=e^{\int -\frac{1}{x}dx}$
$\Rightarrow I.F=e^{-\log x}$
$\Rightarrow I.F.=\frac{1}{x}$
Solution is given by
$y\frac{1}{x}=\int \frac{1}{x^2(1+x^2)}dx$ .....(1)
Resolving $\frac{1}{x^2(1+x^2)}$ into partial fractions
$\frac{1}{x^2(1+x^2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{(1+x^2)}$
$\Rightarrow 1=Ax(1+x^2)+B(1+x^2)+(Cx+D)x^2$
On comparing , we get
$B=1$
$A+C=0$
$B+D=0$
$A=0$
So, we get
$A=C=0,B=1,D=-1$
$\Rightarrow \frac{1}{x^2(1+x^2)}=\frac{1}{x^2}+\frac{-1}{(1+x^2)}$
$\Rightarrow \int \frac{1}{x^2(1+x^2)}dx=\int \frac{1}{x^2}dx-\int \frac{1}{(1+x^2)}dx$
$\Rightarrow \int \frac{1}{x^2(1+x^2)}dx=-\frac{1}{x}-{\tan }^{-1}x+C$
So, by (1),
$\Rightarrow \frac{y}{x}=-\frac{1}{x}-{\tan }^{-1}x+C$