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

The Correct Option is A

$\frac{dy}{dx}-\frac{y}{x}=y^3\left(1+{\log }_{e}x\right)$
$\frac{1}{y^3}\frac{dy}{dx}-\frac{1}{x{y}^2}=1+{\log }_{e}x$
$\text{Let}-\frac{1}{y^2}=t\Rightarrow \frac{2}{y^3}\frac{dy}{dx}=\frac{dt}{dx}$
$\therefore \frac{dt}{dx}+\frac{2t}{x}=2\left(1+{\log }_{e}x\right)$
$\text{I.F.}=e^{\int \frac{2}{x}dx}=x^2$
$\frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+{\log }_{e}x\right)x^3-\frac{x^3}{3}\right)+C$
$y(1)=3$
$\frac{y^2}{9}=\frac{x^2}{5-2x^3\left(2+{\log }_e{x}^3\right)}$
OR
$xdy=ydx+x^3\left(1+{\log }_{e}x\right)dx$
$\frac{xdy-ydx}{y^3}=x\left(1+{\log }_{e}x\right)dx$
$-\frac{x}{y}d\left(\frac{x}{y}\right)=x^2\left(1+{\log }_{e}x\right)dx$
$-{\left(\frac{x}{y}\right)}^2=2\int x^2\left(1+{\log }_{e}x\right)dx$

Answer 2

1. The given differential equation is: \[ \frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \log x)). \] 2. Simplify and separate variables: \[ \frac{dy}{y} = \frac{(1 + xy^2(1 + \log x))}{x} \, dx. \] 3. Rearrange terms: \[ \frac{dy}{y} = (1 + \log x + xy^2) \, dx. \] 4. Multiply by an integrating factor to linearize: \[ I.F = e^{\int \frac{1}{x} dx} = e^{\log x} = x. \] 5. The equation becomes: \[ \frac{d}{dx}(y \cdot x) = x(1 + x^2(1 + \log x)). \] 6. Integrate both sides: \[ y \cdot x = \int x + x^3(1 + \log x) \, dx. \] First term: \[ \int x \, dx = \frac{x^2}{2}. \] Second term: \[ \int x^3 \, dx = \frac{x^4}{4}, \quad \int x^3 \log x \, dx = \frac{x^4}{16}(4\log x - 1). \] 7. Combine and simplify: \[ y \cdot x = \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^4}{16}(4\log x - 1). \] 8. Solve for \(y\) and apply the initial condition \(y(1) = 3\): \[ y = \frac{x^2}{5 - 2x^3(2 + \log x^3)}. \] Thus: \[ \frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2 + \log x^3)}. \] To solve differential equations involving \(\frac{dy}{dx}\), separate variables, integrate both sides, and simplify step by step while applying the given initial condition.