(4)

We have ${\int }_0^x\sqrt{1 - (y\text{'}{\left(t\right))}^2} dt = {\int }_0^{x}y\left(t\right) dt$

On differentiating both sides, we get

$\sqrt{1 - (y\text{'}{\left(x\right))}^2} = y\left(x\right)$

$\Rightarrow 1 - {(y\text{'})}^2 = y^2 \Rightarrow y^2 + {(y\text{'})}^2 = 1$

$\Rightarrow 2yy\text{'} + 2y" = 0 \Rightarrow yy\text{'} + y" = 0$

So, at x = 2, y" + y + 1 = 0 + 1 = 1.

(1)

We have, $\frac{dy}{dx} = 2x{(x + y)}^3 - x(x +\hspace{0.17em}y) - 1, y\left(0\right) = 1$

Let $x + y = t \Rightarrow 1 + \frac{dy}{dx} = \frac{dt}{dx}$

Now, $\frac{dt}{dx} - 1 = 2x{t}^3 - xt - 1$

$\Rightarrow \frac{dt}{dx} = 2x{t}^3 - xt \Rightarrow \frac{1}{t^3} \frac{dt}{dx} +\frac{x}{t^2} - 2x = 0$

Put $\frac{1}{t^2} = u \Rightarrow \frac{-2}{t^3} \frac{dt}{dx} = \frac{du}{dx}$ we get

$\frac{-1}{2} \frac{du}{dx} + xu = 2x \Rightarrow \frac{du}{dx} - 2xu = -4x$

$\therefore \mathrm{IF} = e^{\int -2xdx = e^{-x^2}}$

Required solution = $u·e^{-x^2} = \int e^{-x^2}· (-4x) dx$

$\Rightarrow \frac{e^{-x^2}}{t^2} = \int e^{-x^2}· (-4x) dx$

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

At  x = 0, y = 1, we get $1 =\hspace{0.17em}2 + C \Rightarrow C = -1$

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

At $x = \frac{1}{\sqrt{2}}, {(y + \frac{1}{\sqrt{2}})}^2 = \frac{1}{2 - \sqrt{e}}$

$\therefore (y{\left(\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}})}^2 = (\frac{1}{2 - \sqrt{e}})$.

*

We have, $f\text{'}\left(x\right) = \alpha f\left(x\right) + 3 \Rightarrow f\text{'}\left(x\right) - \alpha f\left(x\right) = 3$

Now, Let $\alpha > 0, \mathrm{IF} = e^{-\alpha \int dx} = e^{-\alpha x}$

$\therefore$  Solution is given by, $e^{-\alpha x}f\left(x\right) = 3\int e^{-\alpha x}dx + C$

$\Rightarrow f\left(x\right) = \frac{-3}{\alpha } + C{e}^{\alpha x}$

Now, let $\alpha > 0$, we have $\underset{x \to -\infty }{\lim }f\left(x\right) = 1$

$\Rightarrow \frac{-3}{\alpha } + C \underset{x \to -\infty }{\lim }{e}^{\alpha x} = 1$

Let $\alpha < 0, \underset{x \to \infty }{\lim }f\left(x\right) = -\frac{3}{\alpha } + C \underset{x \to -\infty }{\lim }{e}^{\alpha x} = 1$

$\Rightarrow \frac{-3}{\alpha } + \infty = 1$ which is not possible

$\therefore$  Value of $\alpha$ does not exist.

(1)

We have, $\frac{dx}{dt} + ax = 0$

$\int \frac{dx}{ax} = -\int dt \Rightarrow \frac{1}{a} \log x = -t + c_1$

$\because$  x(0) = 2

$c_1 = \frac{1}{a} \log 2 \Rightarrow \frac{1}{a} \log x = -t + \frac{1}{a} \log 2 \Rightarrow x = 2e^{-at}$

Now, $\frac{dy}{dt} +by = 0 \Rightarrow \frac{1}{b} \log y = -t + c_2$

$\because y\left(0\right) = 1$

$\therefore \frac{1}{b} \log \left(1\right) = -0 + c_2 \Rightarrow c_2 = 0 \Rightarrow y = e^{-bt}$

Now, $3y\left(1\right) = 2x\left(1\right) \Rightarrow 3e^{-b} = 4e^{-a}$

$\Rightarrow e^{a - b} = \frac{4}{3} \Rightarrow a - b = \log \left(\frac{4}{3}\right)$       ... (i)

For $x\left(t\right) = y\left(t\right), 2e^{-at} = e^{-bt} \Rightarrow e^{(a - b)t} = 2 \Rightarrow (a - b)t = \log 2$

$\Rightarrow \log \left(\frac{4}{3}\right) t = \log 2$      [From (i)]

$\Rightarrow t = \frac{\log 2}{\log {\displaystyle \frac{4}{3}}} \Rightarrow t = {\log }_{\frac{4}{3}} 2$

(1)

Given differential equation is

$(x^2 - 4) dy - (y^2 - 3y) dx = 0 \Rightarrow \frac{dy}{y^2 - 3y} = \frac{dx}{x^2 - 4}$

Integrating on both sides, we get

$\int \frac{dy}{y^2 - 3y} = \int \frac{dx}{x^2 - 4}$

$\Rightarrow \frac{1}{3}\int \frac{y - (y - 3)}{y(y - 3)} dy = \frac{1}{4}\int \frac{(x + 2) - (x - 2)}{(x - 2)(x + 2)} dx$

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

$\Rightarrow \frac{1}{3} \ln \left|\frac{y - 3}{y}\right| = \frac{1}{4} \ln \left|\frac{x - 2}{x + 2}\right| + \ln C \Rightarrow \frac{y -3}{y} = \frac{C_1{(x - 2)}^{3/4}}{{(x + 2)}^{3/4}}$

$\Rightarrow y = \frac{3{(x + 2)}^{3/4}}{{(x + 2)}^{3/4} - C_1{(x - 2)}^{3/4}}$

Now, $y\left(4\right) = \frac{3}{2} \Rightarrow C_1 = -3^{3/4}$

$\therefore y = \frac{3{(x + 2)}^{3/4}}{{(x + 2)}^{3/4} + {(3x - 6)}^{3/4}}$

So, $y\left(10\right) = \frac{3 \times {12}^{3/4}}{{12}^{3/4} + {24}^{3/4}} = \frac{3}{1 + 2^{3/4}} = \frac{3}{1 + {\left(8\right)}^{1/4}}$

(3)

We have, $y \sin 2x + \sin x - (1 + {\cos }^2 x) \frac{dy}{dx} = 0$

$\Rightarrow \frac{dy}{dx} (1 + {\cos }^2 x) - y \sin 2x = \sin x$

$\Rightarrow \frac{dy}{dx} - \left(\frac{\sin 2x}{1 + {\cos }^2 x}\right) y = \frac{\sin x}{1 + {\cos }^2 x}$

$\mathrm{IF} = e^{-\int \frac{\sin 2x}{1 + {\cos }^2 x} dx} = e^{\log (1 + {\cos }^2 x)} = 1 + {\cos }^2 x$

So, solution is given by

$y·(1 + {\cos }^2 x) = \int \frac{\sin x}{(1 + {\cos }^2 x)} (1 + {\cos }^2 x) dx$

$\Rightarrow y(1 + {\cos }^2 x) = \int \sin xdx \Rightarrow y(1 + {\cos }^2 x) = - \cos x + c$         ... (i)

Now, y(0) = 0

$\Rightarrow 0 = - \cos \left(0\right) + c \Rightarrow c = 1$

From (i), $y (1 + {\cos }^2 x) = - \cos x + 1 \Rightarrow y = \frac{1 - \cos x}{1 + {\cos }^2 x}$

$\therefore y \left(\frac{\mathrm{\pi }}{2}\right) = \frac{1 - 0}{1 + 0} = 1$

(2)

The given differential equation is,

$x \cos \left(\frac{\mathrm{y}}{x}\right) \frac{dy}{dx} = y \cos \left(\frac{y}{x}\right) + x$          ... (i)

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

$\therefore$   (i) becomes,

$x \cos v (v + x \frac{dv}{dx}) = vx·\cos v + x \Rightarrow x \frac{dv}{dx} = \sec v$

$\Rightarrow \int \cos vdv = \int \frac{dx}{x} \Rightarrow \sin v = \log \left|x\right| + c$

$\Rightarrow \sin \frac{y}{x} = \log \left|x\right| + c$

Now, $y\left(1\right) = \frac{\mathrm{\pi }}{3} \Rightarrow \sin \frac{\mathrm{\pi }}{3} = c \Rightarrow c = \frac{\sqrt{3}}{2}$

So, the solution of given differential equation is,

$\sin \left(\frac{y}{x}\right) = \log \left|x\right| + \frac{\sqrt{3}}{2}$

On comparing with given solution, we get

$\alpha = \sqrt{3} \Rightarrow {\alpha }^2 = 3$.

(4)

sec xdy + {2(1 – x) tan x + x(2 – x)} dx = 0

$\Rightarrow \frac{dy}{dx} = \frac{- 2(1 - x) \tan x - x(2 - x)}{\sec x}$

$\Rightarrow \frac{dy}{dx} = - 2(1 - x) \sin x - x \cos x (2 - x)$

$\Rightarrow \int dy = - 2\int (1 - x) \sin xdx - \int \cos x (2x - x^2) dx$

$\Rightarrow y = - 2\int (1 - x) \sin xdx - (2x - x^2) \sin x +\int (2 - 2x) \sin xdx + c$

$\Rightarrow y = - (2x - x^2) \sin x + c$

Now, $y\left(0\right) = 2 \Rightarrow 2 = c$

$\Rightarrow y = - (2x - x^2) \sin x + 2 \Rightarrow y\left(2\right) = 2$

(2)

We have, $\frac{dy}{dx} = \frac{\tan x + y}{\sin x (\sec x - \sin x · \tan x)}$

$= \frac{{\displaystyle \frac{\sin x}{\cos x}} + y}{\sin x ({\displaystyle \frac{1}{\cos x}} - \sin x · {\displaystyle \frac{\sin x}{\cos x}})}$

$= \frac{\sin x + y \cos x}{\sin x (1 - {\sin }^2 x)} = \frac{\sin x + y \cos x}{\sin x · {\cos }^2 x}$

$\frac{dy}{dx}= {\sec }^2 x + y · \frac{1}{\sin x · \cos x}$

Now, $\mathrm{IF} = e^{-\int \frac{1}{\sin x · \cos x} dx} = e^{-2\int \mathrm{cosec} 2x dx}$

$= e^{-2}\left(\frac{\log |\mathrm{cosec} 2x - \cot 2x|}{2}\right) = e^{- \log |\mathrm{cosec} 2x - \cot 2x|}$

$= \frac{1}{|\mathrm{cosec} 2x - \cot 2x|} = \frac{1}{|\tan x|}$

The solution is

$y · \frac{1}{|\tan x|} = \int \frac{1}{|\tan x|} · {\sec }^2 xdx = \log |\tan x| + c$

$y = |\tan x| · \log |\tan x| + c |\tan x|$

At  $x = \frac{\mathrm{\pi }}{4}, y = 2 \Rightarrow 2 = 0 + c \Rightarrow c = 2$

At  $x = \frac{\mathrm{\pi }}{3}$

$y = \sqrt{3}·\log \sqrt{3} + 2\sqrt{3} = \sqrt{3}(\log \sqrt{3} + 2)$.

(1)

$y \frac{dy}{dx} = x ({\log }_{e }x - {\log }_{e }y + 1)$

$\frac{dx}{dy} = \frac{x}{y} (\log x - \log y + 1)$

$\frac{dx}{dy} = \frac{x}{y} (\log \frac{x}{y} + 1) (\because \log x - \log y = \log \frac{x}{y})$

Let $\frac{x}{y} = t \Rightarrow x = ty$

Differentiating w.r.t y

$\Rightarrow \frac{dx}{dy} = t + y \frac{dt}{dy} \Rightarrow t + y \frac{dt}{dy} = t \log t + t$

$\Rightarrow \frac{dt}{t \log t} = \frac{dy}{y} \Rightarrow \log (\log t) = \log y + c$

$\Rightarrow \log (\log \frac{x}{y}) = \log y + c$

at x = e and y = 1 (passing point)

$\Rightarrow \log (\log \frac{e}{1}) = \log 1 + c \Rightarrow c = 0$

$\Rightarrow \log (\log \frac{x}{y}) = \log y + 0$

$\Rightarrow \log (\log \frac{x}{y}) = \log y \Rightarrow \log \left(\frac{x}{y}\right) = y$