(2)

We have, $x(x^2+e^x)dy+\left(e^x\right(x–2)y–x^3)dx=0$

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

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

Where, $P=\frac{e^x(x–2)}{x(x^2+e^x)}$ and $Q=\frac{x^2}{(x^2+e^x)}$

$\mathrm{I}.\mathrm{F}.=e^{\int \frac{e^x(x–2)}{x(x^2+e^x)}dx}=e^{\int \frac{e^x+2x}{e^x+x^2}dx–\int \frac{2}{x}dx}=e^{\ln |e^x+x^2|–2 \ln x}$

$\mathrm{I}.\mathrm{F}.=1+\frac{e^x}{x^2}$

Now, $y·(1+\frac{e^x}{x^2})=\int \frac{x^2}{x^2+e^x}·\frac{x^2+e^x}{x^2}dx+C$

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

It is passing through the point (1, 0), then

C = –1

So, $y(1+\frac{e^x}{x^2})=x–1$

Then $y\left(2\right)=\frac{4}{4+e^2}$

(4)

We have, $(x^2+1)\frac{dy}{dx}–2xy=(x^4+2x^2+1) \cos x$

$\Rightarrow \frac{dy}{dx}–\left(\frac{2x}{x^2+1}\right)y=\frac{{(x^2+1)}^2\cos x}{(x^2+1)}=(x^2+1) \cos x$          (Linear D.E.)

Here, $P=\frac{–2x}{(x^2+1)}, Q=(x^2+1) \cos x$

$\mathrm{I}.\mathrm{F}.=e^{\int Pdx}=e^{\int \frac{–2x}{(x^2+1)}dx}=\frac{1}{(x^2+1)}$

Thus, the solution of linear D.E. is given by

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

$\Rightarrow \frac{y}{x^2+1}=\sin x+c \Rightarrow \frac{1}{1}=0+c \Rightarrow c=1$          [$\because$ y(0) = 1]

$\therefore y=(x^2+1)(\sin x+1)$

${\int }_{–3}^{3}y\left(x\right)dx={\int }_{–3}^3(x^2+1)(\sin x+1)dx$

$={\int }_{–3}^3(x^2\sin x+x^2+\sin x+1)dx$

${\int }_{–3}^3{x}^2\sin xdx+{\int }_{–3}^3{x}^{2}dx+{\int }_{–3}^3\sin xdx+{\int }_{–3}^{3}1 dx$

= 0 + 18 + 0 + 6 = 24

(1)

f(x) = x – 1 and $g\left(x\right)=e^x$

f(f(x)) = f(x – 1) = (x – 1) – 1 = x – 2

$g\left(f\right(f\left(x\right)\left)\right)=g(x–2)=e^{x–2}$

Now, $\frac{dy}{dx}+\frac{y}{\sqrt{x}}=e^{–2\sqrt{x}}{e}^{x–2} \Rightarrow \frac{dy}{dx}+\frac{y}{\sqrt{x}}=e^{x–2–2\sqrt{x}}$

Which is a linear differential equation

$\therefore \mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{\sqrt{x}}dx}=e^{2\sqrt{x}}$

$\therefore$   Solution of differential equation is given by

$y{e}^{2\sqrt{x}}=\int e^{x–2–2\sqrt{x}}{e}^{2\sqrt{x}}dx$

$\Rightarrow y{e}^{2\sqrt{x}}=\int e^{x–2}dx \Rightarrow y{e}^{2\sqrt{x}}=e^{x–2}+C$

Now, y(0) = 0

$\Rightarrow 0=e^{–2}+C \Rightarrow C=–e^{–2}$

$\therefore y\left(x\right)=e^{x–2–2\sqrt{x}}–e^{–2–2\sqrt{x}}$

Now, $y\left(1\right)=e^{1–2–2}–e^{–2–2}=e^{–3}–e^{–4}=\frac{e–1}{e^4}$

(4)

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

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

which is a linear D.E. with $P=\frac{1}{y^2}$ and $Q=\frac{1}{y^3}$

$\mathrm{I}.\mathrm{F}.=e^{\int Pdy}=e^{\int \frac{1}{y^2}dy}=e^{\frac{–1}{y}}$

$\therefore$   Solution is given by $x·e^{–1/y}=\int \frac{1}{y^3}·e^{–1/y}dy+c$

Putting $\frac{1}{y}=t \Rightarrow \frac{–1}{y^2}dy=dt$

$\Rightarrow x·e^{\frac{–1}{y}}=–\int t·e^{–t}dy+c \Rightarrow x·e^{\frac{–1}{y}}=–e^{–t}(–t–1)+c$

$\Rightarrow x·e^{\frac{–1}{y}}=e^{\frac{–1}{y}}(1+\frac{1}{y})+c \Rightarrow x=1+\frac{1}{y}+c{e}^{1/y}$

Now, at x(1) = 1, i.e., when x = 1, y = 1

$\Rightarrow 1=1+1+ce \Rightarrow c=\frac{–1}{e}$

$\therefore$   Solution is, $x=1+\frac{1}{y}–\frac{1}{e}\left(e^{1/y}\right)$

$\therefore x\left(\frac{1}{2}\right)=1+2–\frac{1}{e}\times e^2=3–e$

(1)

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{{\tan }^{–1}y}}{1+y^2}$, which is linear differential equation of first order.

$\mathrm{I}.\mathrm{F}.=e^{\int \frac{dy}{1+y^2}}=e^{{\tan }^{—1}y}$

Thus, the required solution is given by

$x{e}^{{\tan }^{—1}y}=\int \frac{2{\left(e^{{\tan }^{—1}y}\right)}^2}{1+y^2}dy$

Put ${\tan }^{—1}y=t, \frac{dy}{1+y^2}=dt$          $\therefore x{e}^{{\tan }^{–1}y}=e^{2 {\tan }^{–1}y}+c$

Given, f(0) = 1, we get 1 = 1 + c $\Rightarrow$ c = 0

Therefore, $x=e^{{\tan }^{–1}y}$

$\Rightarrow f\left(\frac{1}{\sqrt{3}}\right)=e^{\frac{\pi }{6}}$

(3)

Given, $2(3+y)e^{2x}dx–(7+e^{2x})dy=0$

$\Rightarrow \frac{dy}{dx}=\frac{2(3+y)e^{2x}}{(7+e^{2x})}$

$\Rightarrow \frac{dy}{dx}–\frac{2e^{2x}}{(7+e^{2x})}\times y=\frac{6e^{2x}}{(7+e^{2x})}$

$\therefore \mathrm{I}.\mathrm{F}.=e^{–\int \left(\frac{2e^{2x}}{7+e^{2x}}\right)dx}=\frac{1}{7+e^{2x}}$

$\therefore y\times \frac{1}{7+e^{2x}}=\int \frac{6e^{2x}}{{(7+e^{2x})}^2}dx \Rightarrow \frac{y}{7+e^{2x}}=\frac{–3}{(7+e^{2x})}+c$

Since, f(x) pass through the points (0, 5) and $({\log }_{e}2,k)$.

Then, $\frac{5}{8}=\frac{–3}{8}+c \Rightarrow c=1$

$\therefore y=–3+(7+e^{2x}) \Rightarrow y=e^{2x}+4$

$\therefore k=e^{2{\log }_{e}2}+4$

$\Rightarrow k=e^{{\log }_e{2}^2}+4 \Rightarrow k=4+4=8$

(1)

We have, $ydy=(xdy–ydx)\sin \left(\frac{x}{y}\right)$

$\Rightarrow \frac{dy}{y}=\left(\frac{xdy–ydx}{y^2}\right)\sin \left(\frac{x}{y}\right)$

$\Rightarrow \frac{dy}{y}=\sin \left(\frac{x}{y}\right)d\left(\frac{x}{y}\right)$

$\Rightarrow \log y=\cos \frac{x}{y}+C$

$\Rightarrow 0=\cos \frac{\pi }{2}+C$          $[\because x=\frac{\pi }{2} \mathrm{and} y=1]$

$\Rightarrow C=0$

$\therefore \log y=\cos \frac{x}{y}$

Put y = 2 in equation (i), we get

$\Rightarrow \cos \frac{x}{2}=\log 2$

Now, $\cos x=2 {\cos }^2\frac{x}{2}–1=2{\left({\log }_{e}2\right)}^2–1$.

(3)

We have, $(xy–5x^2\sqrt{1+x^2})dx+(1+x^2)dy=0$

$\Rightarrow \frac{dy}{dx}=\frac{5x^2\sqrt{1+x^2}–xy}{(1+x^2)}$

$\Rightarrow \frac{dy}{dx}+\frac{xy}{(1+x^2)}=\frac{5x^2}{\sqrt{1+x^2}}$

$\mathrm{I}.\mathrm{F}.=\int e^{\frac{x}{1+x^2}}dx=\sqrt{1+x^2}$

$y\times \sqrt{1+x^2}=\int \frac{5x^2}{\sqrt{1+x^2}}\times \sqrt{1+x^2}dx+C$

$y\sqrt{1+x^2}=\frac{5x^3}{3}+C$

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

$\therefore C=0$

$\Rightarrow y=\frac{5x^3}{3\sqrt{1+x^2}}$

$\therefore y\left(\sqrt{3}\right)=\frac{5\times 3\times \sqrt{3}}{3\sqrt{1+3}}=\frac{5\sqrt{3}}{2}$

(3)

We have, ${\int }_0^{x}tf\left(x\right)dt=x^{2}f\left(x\right), x>0$

Differentiate the given equation, we get

          $xf\left(x\right)=2xf\left(x\right)+x^{2}f\text{'}\left(x\right)$

$\Rightarrow xf\left(x\right)=–x^{2}f\text{'}\left(x\right) \Rightarrow f\left(x\right)=–xf\text{'}\left(x\right)$

$\Rightarrow \int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=–\int \frac{1}{x}dx \Rightarrow \log f\left(x\right)=–\log x+\log C$

$\Rightarrow \log f\left(x\right)=–\log x + \log C \Rightarrow f\left(x\right)=\frac{C}{x}$

We have, $f\left(2\right)=3; f\left(2\right)=3=\frac{C}{2} \Rightarrow C=6$

So, $f\left(x\right)=\frac{6}{x} \Rightarrow f\left(6\right)=\frac{6}{6}=1$

(3)

We have, ${\int }_0^{2}xF\text{'}\left(x\right)dx=6$          ... (i)

$\Rightarrow {\left[xF\right(x\left)\right]}_0^2–{\int }_0^{2}F\left(x\right)dx=6$

$\Rightarrow 2F\left(2\right)–{\int }_0^{2}F\left(x\right)dx=6$

$\Rightarrow 4–{\int }_0^{2}F\left(x\right)dx=6$           $[\because F(2)=2f(2)=2]$

$\Rightarrow {\int }_0^{2}F\left(x\right)dx=–2$

Also, $\Rightarrow {\int }_0^2{x}^{2}F\text{'}\text{'}\left(x\right)dx=40$

$\Rightarrow {[x^{2}F\text{'}(x\left)\right]}_0^2–2{\int }_0^{2}xF\text{'}\left(x\right)dx=40$

$\Rightarrow 4F\text{'}\left(2\right)–2\left(6\right)=40 \Rightarrow F\text{'}\left(2\right)=13$          [Using (i)]

Thus, $F\text{'}\left(2\right)+{\int }_0^{2}F\left(x\right)dx=13–2=11$