# Linear/Seperable Differential Equations Question (Engineering First Year Calculus)?

## If someone could solve this that would be great, thank you so much :)

Jul 16, 2018

$x y = {e}^{\frac{1}{2} - \frac{1}{2 {x}^{2}}}$.

#### Explanation:

${x}^{3} y ' = y - {x}^{2} y = y \left(1 - {x}^{2}\right)$.

$\therefore \frac{y '}{y} = \frac{1 - {x}^{2}}{x} ^ 3 ,$

$i . e . , \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} ^ 3 - {x}^{2} / {x}^{3} ,$

$\mathmr{and} , \frac{\mathrm{dy}}{y} = \left(\frac{1}{x} ^ 3 - \frac{1}{x}\right) \mathrm{dx} \ldots \ldots \ldots \ldots . \text{[Separable Variable]}$.

Integrating, $\int \frac{\mathrm{dy}}{y} = \int \left(\frac{1}{x} ^ 3 - \frac{1}{x}\right) \mathrm{dx} - \ln c$,

$\therefore \ln y = {x}^{-} \frac{2}{-} 2 - \ln x - \ln c ,$

$i . e . , \ln y + \ln x + \ln c = - \frac{1}{2 {x}^{2}} ,$

$\mathmr{and} , \ln \left(y x c\right) = - \frac{1}{2 {x}^{2}}$.

$\therefore y x c = {e}^{- \frac{1}{2 {x}^{2}}} \ldots \ldots \left(\ast\right)$, is a desired Gen. Soln.

To get the Particular Soln., let us use the Initial Cond. (IC)

$y \left(- 1\right) = - 1 , \text{ i.e., when } x = - 1 , y = - 1$.

Using IC in $\left(\ast\right) , \left(- 1\right) \left(- 1\right) c = {e}^{- \frac{1}{2}} \Rightarrow c = {e}^{- \frac{1}{2}}$.

Subst.ing in the GS, we get, $y x {e}^{- \frac{1}{2}} = {e}^{- \frac{1}{2 {x}^{2}}}$

$\mathmr{and} , x y = {e}^{\frac{1}{2} - \frac{1}{2 {x}^{2}}}$.