Question

The solution of $\frac{dy}{dx}+3x^{2}y=x^5{e}^{x^3}$ is

• A. $12x=e^{x^3}(2x^3-1)+{ce}^{{-x}^3}$
• B. $12x=e^{x^3}(12x^3+1)+{ce}^{{-x}^3}$
• C. $12y=e^{x^3}(2x^3-1)+{ce}^{{-x}^3}$
• D. $12y=e^{x^3}(2x^3+1)+{ce}^{{-x}^3}$

Answer 1

The correct option is A $12x=e^{x^3}(2x^3-1)+{ce}^{{-x}^3}$

It is a first order linear differential equation

Here,

$P=3x^2,Q=x^5{e}^{x^3}$

$I.F.=e^{\int Pdx}=e^{x^3}$

Solution of D.E.

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

$y{e}^{x^3}=\int x^5{e}^{x^3}{e}^{x^3}dx$

$=\int x^3{e}^{2x^3}{x}^{2}dx$

put $2x^3=t,6x^{2}dx=dt$

$=\frac{1}{12}\int t{e}^{t}dt+C$

integrating by parts taking $t$ as first function

$=\frac{1}{12}\int [t{e}^t-\int 1.e^{t}dt]+C$

$y{e}^{x^3}=\frac{1}{12}{e}^t(t-1)+C$

Put $t=2x^3$

$12y=e^{x^3}(2x^3-1)+C{e}^{-x^3}$