Let y = y(x) be the solution curve of the differential equation [dy dx = y x( 1 + xy^2(1 +…

Question

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

Accepted Answer

$$
\begin{aligned}
& \frac{d y}{d x}-\frac{y}{x}=y^3\left(1+\log _e x\right) \\
& \frac{1}{y^3} \frac{d y}{d x}-\frac{1}{x y^2}=1+\log _e x \\
& \text { Let }-\frac{1}{y^2}=t \Rightarrow \frac{2}{y^3} \frac{d y}{d x}=\frac{d t}{d x} \\
& \therefore \frac{d t}{d x}+\frac{2 t}{x}=2\left(1+\log _e x\right) \\
& \text { I.F. }=e^{\int \frac{2}{x} d x}=x^2 \\
& \frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+\log _e x\right) x^3-\frac{x^3}{3}\right)+C \\
& y(1)=3 \\
& \frac{y^2}{9}=\frac{C}{5-2 x^3\left(2+\log _e x^3\right)}
\end{aligned}
$$

Solutions · 1