If x ^ 3 dy + xy · dx = x ^ 2 dy + 2ydx; y(2) = e and x > 1 , then...

Given ${x}^{3}dy + xy \cdot dx = {x}^{2}dy + 2ydx$

$⇒ \left({x}^{3} - {x}^{2}\right)dy = \left(2 - x\right)ydx$

$⇒ \dfrac{dy}{y} = \dfrac{\left(2 - x\right)}{\left({x}^{3} - {x}^{2}\right)}dx$

Integrating both sides with respect to $x$, we get

$∫ \dfrac{dy}{y} = ∫ \dfrac{\left(2 - x\right)}{{x}^{2}\left(x - 1\right)}dx + k . . . . . . . . . . . . \left(i\right)$

Let $\dfrac{\left(2 - x\right)}{{x}^{2}\left(x - 1\right)} = \dfrac{A}{x} + \dfrac{B}{{x}^{2}} + \dfrac{C}{\left(x - 1\right)}$

$⇒ \left(2 - x\right) = Ax\left(x - 1\right) + B\left(x - 1\right) + C{x}^{2}$

Putting $x = 0 ⇒ 2 = - B ⇒ B = - 2$

Putting $x = 1 ⇒ 2 - 1 = C ⇒ C = 1$

Putting $x = 2 ⇒ 2 - 2 = A\left(2\right)\left(1\right) + B\left(1\right) + C\left({2}^{2}\right) ⇒ 2A + 2 = 0 ⇒ A = - 1$

From equation \left(i\right)$, we get</p><p>$∫ \dfrac{dy}{y} = ∫ \left(\dfrac{- 1}{x} + \dfrac{- 2}{{x}^{2}} + \dfrac{1}{x - 1}\right)dx + k$</p><p>$⇒ \ln y = - \ln x + \dfrac{2}{x} + \ln \left|x - 1\right| + k . . . . . . . . . . \left(ii\right)$</p><p>Given$y\left(2\right) = e$i.e. at$x = 2, y = e$</p><p>$⇒ lne = - \ln 2 + \dfrac{2}{2} + \ln \left|2 - 1\right| + k$</p><p>$⇒ 1 = - \ln 2 + 1 + 0 + k$</p><p>$⇒ k = \ln 2$</p><p>Putting in equation \left(ii\right)$, we get

$\ln y = - \ln x + \dfrac{2}{x} + \ln \left|x - 1\right| + \ln 2$

Now putting $x = 4$, we get

$\ln y = - \ln 4 + \dfrac{2}{4} + \ln \left|4 - 1\right| + \ln 2$

$⇒ \ln y = - 2\ln 2 + \dfrac{1}{2} + \ln 3 + \ln 2$

$⇒ \ln y = - \ln 2 + \dfrac{1}{2} + \ln 3$

$⇒ \ln y = \dfrac{1}{2} + ln\dfrac{3}{2}$

$⇒ ln\dfrac{2y}{3} = \dfrac{1}{2}$

$⇒ \dfrac{2y}{3} = e\dfrac{1}{2}$

$⇒ y = \dfrac{3}{2}\sqrt{e}$

$⇒ y\left(4\right) = \dfrac{3}{2}\sqrt{e}$