$$y + {{xdy} \over {dx}} = {x^2}$$ (given)

$$ \Rightarrow {{dy} \over {dx}} + {y \over x} = x$$

If $${e^{\int {{1 \over x}dx} }} = x$$

Solution of DE

$$ \Rightarrow y\,.\,x = \int {x\,.\,x\,dx} $$

$$ \Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}$$

Passes through ($$-$$2, 2), so

$$-$$12 = $$-$$ 8 + c $$\Rightarrow$$ c = $$-$$ 4

$$\therefore$$ 3xy = x³ $$-$$ 4

i.e. 3x . f(x) = x³ $$-$$ 4

$${{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x$$

IF = $${e^{ - {x^2}}}$$

So, $$y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx} $$

$$ \Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$$

$$ \Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$$

Given at x = 0, y = 7

$$\Rightarrow$$ 7 = 5 + c $$\Rightarrow$$ c = 2

So, $$y = 5 - 2\sin x + 2{e^{{x^2}}}$$

Now, at x = $$\pi$$,

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

$$2{x^2}dy + ({e^y} - 2x)dx = 0$$

$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$

$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} = - {1 \over {2{x^2}}} \Rightarrow $$ Put $${e^{ - y}} = z$$

$${{ - dz} \over {dx}} - {z \over x} = - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}$$

$$d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c$$

$$x{e^{ - y}} = {1 \over 2}{\log _e}x + c$$, passes through (e, 1)

$$ \Rightarrow C = {1 \over 2}$$

$$x{e^{ - y}} = {{{{\log }_e}ex} \over 2}$$

$${e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2$$

$$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$

or $${{dy} \over {dx}} + {{{{\sec }^2}x} \over {\tan x}}.y = - \tan x$$

$$IF = {e^{\int {{{{{\sec }^2}x} \over {\tan x}}dx} }} = {e^{\ln \tan x}} = \tan x$$

$$\therefore$$ $$y\tan x = - \int {{{\tan }^2}x\,dx} $$

or $$y\tan x = - \tan x + x + C$$

or $$y = - 1 + {x \over {\tan x}} + {C \over {\tan x}}$$

or $$\mathop {\lim }\limits_{x \to 0} xy = - x + {{{x^2}} \over {\tan x}} + {{Cx} \over {\tan x}} = 1$$

or C = 1

$$y(x) = \cot x + x\cot x - 1$$

$$y\left( {{\pi \over 4}} \right) = {\pi \over 4}$$