Maths Solve the Following Question.(3 Marks)

$4 \frac{d x}{d y}+8 x=5 e^{-3 y}$

$\frac{d y}{d x}+ y \cot x = x ^2 \cot x +2 x$

the hyperbola whose length of transverse and conjugate axes are half of that of the

given hyperbola $\frac{x^2}{16}-\frac{y^2}{36}=k$.

$y=A e^{3 x+1}+B e^{-3 x+1}$

$xy = ae ^{ x }+ be ^{- x }+ x ^2 ; x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}+x^2=x y+2$

tangent to the curve at any point $(x, y)$ is $-\frac{4 x}{9 y}$.

$\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$

$y d x+\left(x-y^2\right) d y=0$

$(x+a) \frac{d y}{d x}-3 y=(x+a)^5$

$\frac{d y}{d x}+y \cdot \sec x=\tan x$

$\left( x +2 y ^3\right) \frac{d y}{d x}= y$

$\cos ^2 x \cdot \frac{d y}{d x}+ y =\tan x$

$x ^2 \frac{d y}{d x}= x ^2+ xy + y ^2$

$x d y+2 y \cdot d x=0$, when $x=2, y=1$

$x \sin \left(\frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x$

$(x-y)^2 \frac{d y}{d x}=a^2$

$\frac{d y}{d x}=\cos ( x + y )$

$(x+1) \frac{d y}{d x}-1=2 e^{-y}, y=0$, when $x=1$

$y (1+\log x ) \frac{d x}{d y}- x \log x =0, y = e ^2$, when $x = e$

$\left(x-y^2 x\right) d x-\left(y+x^2 y\right) d y=0$, when $x=2, y=0$

$\frac{\cos ^2 y}{x} d y+\frac{\cos ^2 x}{y} d x=0$

$y=\left(\sin ^{-1} x\right)^2+c_{;}\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=2$

$y=e^{-2 x}(A \cos x+B \sin x)$

$c_1 x^3+c_2 y^2=5$

$y=c_1 e^{2 x}+c_2 e^{5 x}$