The solution of x3dydx+4x2tan⁡y=exsec⁡y satisfying y(1)=0 is:

x3dydx+4x2tan⁡y=exsec⁡yRewriting the given equation in the form x4cos⁡ydydx+4x3sin⁡y=xex⇒ddxx4sin⁡y=xex⇒x4sin⁡y=∫xexdx+c=(x−1)ex+cSince, y(1)=0 so, c=0Hence, sin⁡y=x−4(x−1)ex

The solution of x3dydx+4x2tan⁡y=exsec⁡y satisfying y(1)=0 is:

x3dydx+4x2tan⁡y=exsec⁡yRewriting the given equation in the form x4cos⁡ydydx+4x3sin⁡y=xex⇒ddxx4sin⁡y=xex⇒x4sin⁡y=∫xexdx+c=(x−1)ex+cSince, y(1)=0 so, c=0Hence, sin⁡y=x−4(x−1)ex

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The solution of $x^{3} \frac{d y}{d x} + 4 x^{2} \tan y = e^{x} \sec y$ satisfying $y (1) = 0$ is:

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$\tan y = (x - 2) e^{x} \log x$

$\sin y = e^{x} (x - 1) x^{- 4}$

$\tan y = (x - 1) e^{x} x^{- 3}$

$\sin y = e^{x} (x - 1) x^{- 3}$

answer is B.

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Detailed Solution

$x^{3} \frac{d y}{d x} + 4 x^{2} \tan y = e^{x} \sec y$

Rewriting the given equation in the form $x^{4} \cos y \frac{d y}{d x} + 4 x^{3} \sin y = x e^{x}$

$\begin{array}{l} \Rightarrow \frac{d}{d x} (x^{4} \sin y) = x e^{x} \\ \Rightarrow x^{4} \sin y = \int x e^{x} d x + c = (x - 1) e^{x} + c \end{array}$