Question

The solution of the differential equation $\frac{d y}{d x}+xyln y=x^{3}y$ is equal to (where, $C$ is the constant of integration)

• A. $ln y=x^{2}+Ce^{- x^{2}}$
• B. $ln y=x^{2}-2+Ce^{- x^{2}}$
• C. $ln y=x^{2}-2+Ce^{- \frac{x^{2}}{2}}$
• D. $ln y=x^{2}+Ce^{- \frac{x^{2}}{2}}$

Answer 1

Answer: (C)

$\frac{1}{y}\frac{d y}{d x}+xln y=x^{3}$
Let $ln y=t$
$\frac{1}{y}\frac{d y}{d x}=\frac{d t}{d x}$
$\frac{d t}{d x}+tx=x^{3}$
I.F. $=e^{\displaystyle \int x d x}=e^{\frac{x^{2}}{2}}$
$te^{\frac{x^{2}}{2}}=\displaystyle \int e^{\frac{x^{2}}{2}}x^{3}dx$
$ln ye^{\frac{x^{2}}{2}}=\displaystyle \int \left(e^{\frac{x^{2}}{2}} \cdot x\right)x^{2}dx$
$ln ye^{\frac{x^{2}}{2}}=e^{\frac{x^{2}}{2}}\cdot x^{2}-\displaystyle \int \left(2 x\right)e^{\frac{x^{2}}{2}}dx$
$ln e^{\frac{x^{2}}{2}}=e^{\frac{x^{2}}{2}}x^{2}-2e^{\frac{x^{2}}{2}}+C$
$ln y\cdot e^{\frac{x^{2}}{2}}=e^{\frac{x^{2}}{2}}\left(x^{2} - 2\right)+C$
$ln y=\left(x^{2} - 2\right)+Ce^{\frac{- x^{2}}{2}}$