(1)

$(x^4 + 2x^3 + 3x^2 + 2x + 2)dy - (2x^2 + 2x + 3)dx = 0$

$\frac{dy}{dx} = \frac{2x^2 + 2x + 3}{x^4 + 2x^3 + 3x^2 + 2x + 2}$

$\Rightarrow \int dy = \int \left(\frac{2x^2 + 2x + 3}{x^4 + 2x^3 + 3x^2 + 2x + 2}\right) dx$

$\Rightarrow \int dy = \int \frac{2x^2 + 2x + 3}{(x^2 + 1)(x^2 + 2x + 2)} dx$

$\Rightarrow \int dy = \int \frac{1}{x^2 + 1} dx + \int \frac{1}{{(x + 1)}^2 + 1} dx$

$\Rightarrow y = {\tan }^{-1}\left(x\right) + {\tan }^{-1}(1 + x) + C$        ... (i)

Now, $y(-1) = \frac{-\mathrm{\pi }}{4}$

$\therefore$ From (i), we get $\frac{-\mathrm{\pi }}{4} = {\tan }^{-1}(-1) + {\tan }^{-1}(1 - 1) + C$

$\Rightarrow \frac{-\mathrm{\pi }}{4} = - \frac{\mathrm{\pi }}{4} + C \Rightarrow C = 0$

So, $y\left(x\right) = {\tan }^{-1}\left(x\right) + {\tan }^{-1}(1 + x)$

Now, $y\left(0\right) = {\tan }^{-1}\left(0\right) + {\tan }^{-1}(1 + 0) = \frac{\mathrm{\pi }}{4}$.

(4)

We Have, ${(x^2 + 4)}^{2}dy + (2x^{3}y + 8xy - 2)dx = 0$

$\Rightarrow \frac{dy}{dx} + \frac{y(2x^3 + 8x)}{{(x^2 + 4)}^2} = \frac{2}{{(x^2 + 4)}^2}$

$\Rightarrow \frac{dy}{dx} + \frac{2xy}{x^2 + 4} = \frac{2}{{(x^2 + 4)}^2}$

$IF = e^{\int \frac{2x}{x^2 + 4} dx} = e^{{\log }_e (x^2 + 4)} = x^2 + 4$

$\therefore$ Solution is $y(x^2 + 4) = \int \frac{2}{{(x^2 + 4)}^2} \times (x^2 + 4) dx + C$

$\Rightarrow y(x^2 + 4) = \int \frac{2}{x^2+ 4} dx + C$

$\Rightarrow y(x^2 + 4) = 2 \times \frac{1}{2} {\tan }^{-1}\left(\frac{x}{2}\right) + C$

$\Rightarrow y(x^2 + 4) = {\tan }^{-1}\left(\frac{x}{2}\right) + C$                   ... (i)

Now. We have y(0) = 0

$\Rightarrow 0·(0 + 4) = {\tan }^{-1}\left(0\right) + C \Rightarrow C = 0$

$\therefore y(x^2 + 4) = {\tan }^{-1}\frac{x}{2}$

So, at x = 2

$y(4 + 4) = {\tan }^{-1} \left(\frac{2}{2}\right) \Rightarrow y = \frac{1}{8} \times {\tan }^{-1} 1 = \frac{1}{8} \times \frac{\mathrm{\pi }}{4}$

$\Rightarrow y = \frac{\mathrm{\pi }}{32}$

(1)

$\frac{dy}{dx} + 2y = \sin 2x$, $y\left(0\right) = \frac{3}{4}$

$\mathrm{IF} = e^{\int 2dx} = e^{2x}$

$\therefore$ General solution of given differential equation is,

$y \times e^{2x} = \int e^{2x} \sin 2x dx$

$$\begin{gathered} \Rightarrow y{e}^{2x} = \frac{e^{2x}}{8} (2 \sin 2x - 2 \cos 2x) + c \\ [\because \int e^{ax} \sin bx dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + c] \end{gathered}$$

$\Rightarrow y{e}^{2x} = \frac{e^{2x}}{4} (\sin 2x - \cos 2x) + c$

Now, $y\left(0\right) = \frac{3}{4}$

$\Rightarrow \frac{3}{4} = \frac{1}{4} (-1) + c \Rightarrow c = \frac{3}{4} + \frac{1}{4} = 1$

$\therefore y\left(\frac{\mathrm{\pi }}{8}\right) = \frac{1}{4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) + \frac{1}{e^{\mathrm{\pi }/4}} = e^{-\mathrm{\pi }/4}$

(2)

We Have, $(1 + x^2) \frac{dy}{dx} + y = e^{{\tan }^{-1}x}$

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

$\mathrm{IF} = e^{\int \frac{1}{1 + x^2} dx} = e^{{\tan }^{-1} x}$

$\therefore y{e}^{{\tan }^{-1} x} = \int \frac{e^{{\tan }^{-1} x}}{1 + x^2}·e^{{\tan }^{-1} x} dx + c$

Let ${\tan }^{-1} x = t$

$\Rightarrow \frac{1}{1 + x^2} dx = dt \Rightarrow y{e}^t = \int e^{2t} dt + c$

$\Rightarrow y{e}^t = \frac{e^{2t}}{2} + c \Rightarrow y{e}^{{\tan }^{-1} x} = \frac{e^{2 {\tan }^{-1} x}}{2} + c$

Since y(1) = 0

$\Rightarrow 0e^{\mathrm{\pi }/4} = \frac{e^{\mathrm{\pi }/2}}{2} + c \Rightarrow c = \frac{-e^{\mathrm{\pi }/2}}{2}$

$\therefore y = \frac{e^{{\tan }^{-1} x}}{2} - \frac{e^{\mathrm{\pi }/2}}{2e^{{\tan }^{-1} x}}$

$\Rightarrow y\left(0\right) = \frac{1}{2} - \frac{e^{\mathrm{\pi }/2}}{2} = \frac{1 - e^{\mathrm{\pi }/2}}{2}$

(3)

We Have, $\frac{dy}{dx} + \frac{2y}{2x {\log }_e x} = \frac{3 {\log }_e x}{x (2x {\log }_e x)}$

$\Rightarrow \frac{dy}{dx} + \frac{y}{x {\log }_e x} = \frac{3}{2x^2}$

$\mathrm{IF} = e^{\int \frac{1}{x {\log }_e x} dx} = e^{\log ({\log }_e x)} = {\log }_e x$

$\Rightarrow y {\log }_e x = \int \frac{3}{2x^2} {\log }_e x dx$

$= \frac{3}{2} {\log }_e x\int x^{-2} dx - \int (\frac{3}{2x}\int x^{-2} dx) dx$

$= \frac{3}{2} {\log }_e x \left(\frac{-1}{x}\right) + \int \frac{3}{2x^2} dx + c = \frac{-3 {\log }_e x}{2x} + \left(\frac{-3}{2x}\right) + c$

Now, $y\left(e^{-1}\right) = 0$

$\Rightarrow 0 = \frac{3e}{2} - \frac{3e}{2} + c \Rightarrow c = 0$

$\therefore y {\log }_e x = \frac{-3 {\log }_e x}{2x} - \frac{3}{2x} \Rightarrow y = \frac{-3}{2x} - \frac{3}{2x {\log }_e x}$

$\Rightarrow y\left(e\right) = \frac{-3}{2e} - \frac{3}{2e} = \frac{-3}{e}$

(4)

We have, $\frac{dy}{dx} = \frac{(2 + \alpha )x - \beta y + 2}{\beta x - 2\alpha y - (\beta \gamma - 4\alpha )}$

$\Rightarrow \beta xdy - 2\alpha ydy - (\beta \gamma - 4\alpha ) dy = 2xdx + \alpha xdx - \beta ydx + 2dx$

On integrating, we get

$\int \beta xdy + \beta ydx - \alpha y^2 - (\beta \gamma - 4\alpha ) y = x^2 + \frac{\alpha x^2}{2} + 2x + c$

$\Rightarrow \beta xy - \alpha y^2 - (\beta \gamma - 4\alpha ) y = x^2(1 + \frac{\alpha }{2}) + 2x + c$

$\Rightarrow (1 + \frac{\alpha }{2}) x^2 + \alpha y^2 - \beta xy + 2x + (\beta \gamma - 4\alpha ) y + c = 0$

Since, it represents a circle which is passing through origin, then

$1 + \frac{\alpha }{2} = \alpha , \beta = 0 \mathrm{and} c = 0 \left[\begin{array}{c}\because \mathrm{coeff}. \mathrm{of} x^2 = \mathrm{coeff}. \mathrm{of} y^2\\ \mathrm{coeff}. \mathrm{of} xy = 0 \end{array}\right]$

$\Rightarrow \alpha = 2$

$\therefore$ Equation of circle is given by

$2x^2 + 2y^2 + 2x - 8y = 0 \Rightarrow x^2 + y^2 + x - 4y = 0$

$\Rightarrow \mathrm{Centre} \equiv (- \frac{1}{2}, 2)$

$\Rightarrow \mathrm{Radius} = \sqrt{{\left(\frac{1}{2}\right)}^2 + {\left(2\right)}^2 - 0} = \frac{\sqrt{17}}{2}$

(4)

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

$\Rightarrow \frac{e^{\tan x}}{{\cos }^2 x (1 + e^{2 \tan x})} dx + \frac{1}{1 + y^2} dy = 0$

Integrating both sides, we get

$\int \frac{{\sec }^2 x{e}^{\tan x}}{1 + e^{2 \tan x}} dx + \int \frac{dy}{1 + y^2} = C$

Put $e^{\tan x} = t \Rightarrow {\sec }^2 x{e}^{\tan x}dx = dt$

$\Rightarrow {\tan }^{-1}\left(t\right) + {\tan }^{-1}y = C \Rightarrow {\tan }^{-1}\left(e^{\tan x}\right) + {\tan }^{-1}y = C$

Since, y(0) = 1

$\Rightarrow \frac{\mathrm{\pi }}{4} + \frac{\mathrm{\pi }}{4} = C \Rightarrow C = \frac{\mathrm{\pi }}{2}$

So, ${\tan }^{-1} \left(e^{\tan x}\right) + {\tan }^{-1} y = \frac{\mathrm{\pi }}{2}$

For $x = \frac{\mathrm{\pi }}{4}$, we have

${\tan }^{-1} y = \frac{\mathrm{\pi }}{2} - {\tan }^{-1} e$

$\Rightarrow {\tan }^{-1} y = {\cot }^{-1} e \Rightarrow {\tan }^{-1} y = {\tan }^{-1} \frac{1}{e} \Rightarrow y = \frac{1}{e}$

(3)

We have, $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$

$\Rightarrow {\sec }^2 y \frac{dy}{dx} + 2x \tan y = x^3$               ... (i)

Let $z = \tan y \Rightarrow \frac{dz}{dx} = {\sec }^2 y \frac{dy}{dx}$

$\therefore \mathrm{From} \left(\mathrm{i}\right), \frac{dz}{dx} + 2xz = x^3$

$\mathrm{IF} = e^{\int 2xdx} = e^{x^2} \Rightarrow z·e^{x^2} = \int e^{x^2}·x^{3}dx + c$

$\Rightarrow \tan y·e^{x^2} = \frac{1}{2}(x^2{e}^{x^2} - e^{x^2}) + c$

Since, y(1) = 0

$\therefore \tan \left(0\right) · e = \frac{1}{2} (1·e - e) + c \Rightarrow c = 0$

So, $\tan y{e}^{x^2} = \frac{1}{2} (x^2{e}^{x^2} - e^{x^2})$

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

$\therefore y\left(\sqrt{3}\right) = {\tan }^{-1} \left(\frac{{\left(\sqrt{3}\right)}^2 - 1}{2}\right) = {\tan }^{-1}\left(1\right) = \frac{\mathrm{\pi }}{4}$

(4)

We have, $2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx}$

$\Rightarrow$  2ydy + 3dx = 5dy

Integrating both sides, we get

$y^2 + 3x = 5y + c$

Now, at point (0, 1) i.e., at x = 0, y = 1, we have

1 + 0 = 5 + c $\Rightarrow$ c = –4

So, $y^2 - 5y = -3x - 4$

$\Rightarrow y^2 - 5y + \frac{25}{4} - \frac{25}{4} = -3x - 4$

$\Rightarrow {(y - \frac{5}{2})}^2 = -3x + \frac{9}{4} \Rightarrow {(y - \frac{5}{2})}^2 = -3(x - \frac{3}{4})$

Vertex = $(\frac{3}{4}, \frac{5}{2})$, Which satisfies by 2x + 3y = 9.

(3)

We have, $(x^2 + y^2) dx - 5xydy = 0$

$\frac{dy}{dx} = \frac{x^2 + y^2}{5xy}$

Let $y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$

$\therefore v + x \frac{dy}{dx} = \frac{1 + v^2}{5v} \Rightarrow x \frac{dv}{dx} = \frac{1 + v^2}{5v} - v$

$\Rightarrow \frac{5vdv}{1 - 4v^2} = \frac{dx}{x} \Rightarrow \frac{1}{8}\int \frac{8 \times 5vdv}{1 - 4v^2} = \int \frac{dx}{x}$

$\Rightarrow \frac{-5}{8} \ln |1 - 4v^2| = \ln \left|x\right| + \ln c$

$\Rightarrow \frac{5}{8} \ln \frac{|x^2 - 4y^2|}{x^2} + \ln \left|x\right| + \ln c = 0$

$\Rightarrow |\frac{x^2 - 4y^2}{x^2}{|}^{5/8} |x| = k$

$\Rightarrow \frac{|x^2 - 4y^2{|}^{5/8}}{|x{|}^{{\displaystyle \frac{1}{4}}}} = k \because y\left(1\right) = 0 \Rightarrow k = 1$

$\Rightarrow |\begin{array}{c}{x}^2 - 4y^2\end{array}{|}^{5/8} = |x{|}^{\frac{1}{4}} \Rightarrow |x^2 - 4y^2{|}^5 = x^2$