Question

Let $y=y(x)$ be the solution curve of the differential equation $\frac{d y}{d x}=\frac{y}{x}\left(1+x y^2\left(1+\log _e x\right)\right), x>0, y(1)=3$ Then $\frac{y^2(x)}{9}$ is equal to :

• A. $\frac{x^2}{5-2 x^3\left(2+\log _e x^3\right)}$
• B. $\frac{x^2}{3 x^3\left(1+\log _e x^2\right)-2}$
• C. $\frac{x^2}{7-3 x^3\left(2+\log _e x^2\right)}$
• D. $\frac{x^2}{2 x^3\left(2+\log _e x^3\right)-3}$

Answer 1

Answer: (A)

$\frac{x^2}{5-2 x^3\left(2+\log _e x^3\right)}$

Answer 2

dxdy​−xy​=y3(1+loge​x)
y31​dxdy​−xy21​=1+loge​x
Let −y21​=t⇒y32​dxdy​=dxdt​
∴dxdt​+x2t​=2(1+loge​x)
I.F. =e∫x2​dx=x2
y2−x2​=32​((1+loge​x)x3−3x3​)+C
y(1)=3
9y2​=5−2x3(2+loge​x3)x2​
OR
xdy=ydx+x3(1+loge​x)dx
y3xdy−ydx​=x(1+loge​x)dx
−yx​d(yx​)=x2(1+loge​x)dx
−(yx​)2=2∫x2(1+loge​x)dx