Question

The solution of $\frac{dy}{dx}+\frac{3}{x}\left(y\right)=\frac{1}{x^2}$ given that $y=2;x=1$ is

• A. $2x^{3}y=x^2-3$
• B. $2x^{3}y=x^2+3$
• C. $x^{3}y=x^2+3$
• D. $2x^{3}y=x^2+5$

Answer 1

The correct option is A $2x^{3}y=x^2-3$

It is first order linear D.E.

$P=\frac{3}{x},Q=\frac{1}{x^2}$

I.F. $=e^{\int P.dx}=e^{\int \frac{3}{x}dx}=e^{\log x^3}=x^3$

Solution is $y\times I.F.=\int (Q\times I.F.)dx+C$

$y{x}^3=\int (\frac{1}{x^2}\times x^3)dx+C$

$x^{3}y=\frac{x^2}{2}+C$

$y=2,x=1$

$C=\frac{3}{2}$

Therefore,

$2x^3=x^2+3$