y = (tan x – 1) + Ce^–tanx

\(\tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \log \left(x^2+y^2\right)+\mathrm{C}\)

y′′′ + 2y'' + y′ = 0
The highest order derivative present in the differential equation is y′′′. Therefore, its order is three.
It is a polynomial equation in y′′′, y′′, y′  . The highest power raised to y′′ is 1. Hence, its degree is 1.

\(= 2\ tan \sqrt x\ + c\)

\(\int \frac{\sec ^{2} x}{3+\tan x} d x =\int \frac{1}{t} d t \)
= log |t|+C
= log | 3 + tan x | + C

\(\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x=\frac{1}{6} \int \frac{1}{t} d t=\frac{1}{6} \log |t|+C \)
\(={1\over6}log|3x^2 + sin6x|+C|\)

The graph of y = sin x is shown in the figure below
Clearly, required region is bounded by y = sin x, x-axis and lines x = 0 and x = 2\(\pi\), which is represented by shaded region.
\(\therefore\) Required area = Area of region OABO + Area of region BCDB
\(=\int_{0}^{\pi} \sin x d x+\mid \int_{\pi}^{2 \pi} \sin x d x\) 
[since, in region BCDB, the graph is below the X-axis, so its area comes out to be negative, therefore we take the absolute value]
\(=[-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right|\) 
\(=(-\cos \pi+\cos 0)+|-\cos 2 \pi+\cos \pi|\) 
\(=(1+1)+|-1-1|\) 
\(=2+|-2|=2+2=4 \text { sq units }\) 
Hence, the required area is 4 sq. units.

We have
\(\frac{x^{4}}{(x-1)\left(x^{2}+1\right)} =(x+1)+\frac{1}{x^{3}-x^{2}+x-1} \)
\(=(x+1)+\frac{1}{(x-1)\left(x^{2}+1\right)} \)
\(\text {Now express } \ \frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{\mathrm{A}}{(x-1)}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^{2}+1\right)}\)
1 = A (x² + 1) + (Bx + C) (x – 1)
= (A + B) x² + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give \( \mathrm{A}=\frac{1}{2}, \mathrm{~B}=\mathrm{C}=-\frac{1}{2} \) Substituting values of A,B and C in (2), we get 
\(\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}\)
 Again, substituting (3) in (1), we have 
\(\frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)}=(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}\)
\(\text {Therefore}\ \int \frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)} d x=\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+\mathrm{C}\)

\(\text { Let } \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x\)
Then, by P₄
\(\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x=\int_{0}^{\frac{\pi}{2}} d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}\)
\(\text { Hence } \ \mathrm{I}=\frac{\pi}{4}\)

The given equation can be written as:
\(\left[x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)\right] d y=\left[x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)\right] d x\) ..(1)
 \(\frac{d y}{d x}=\frac{x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)}{x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)}\)
Dividing numerator and denominator on RHS by x², we get
\(\frac{d y}{d x}=\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)+\left(\frac{y^2}{x^2}\right) \sin \left(\frac{y}{x}\right)}{\frac{y}{x} \sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)}\) ... (1)
Clearly, equation (1) is a homogeneous differential equation of the form \(\frac{d y}{d x}=g\left(\frac{y}{x}\right)\)
To solve it, we make the substitution
y = vx
or  \( \frac{d y}{d x} =v+x \frac{d v}{d x} \)
or  \(v+x \frac{d v}{d x} =\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v} \) (using (1) and (2))
\(x \frac{d v}{d x} =\frac{2 v \cos v}{v \sin v-\cos v}\)
\(\left(\frac{v \sin v-\cos v}{v \cos v}\right) d v=\frac{2 d x}{x}\)
Therefore \(\int\left(\frac{v \sin v-\cos v}{v \cos v}\right) d v=2 \int \frac{1}{x} d x\)
or\(\int \tan v d v-\int \frac{1}{v} d v=2 \int \frac{1}{x} d x\)
or \(\log |\sec v|-\log |v|=2 \log |x|+\log \left|C_1\right|\)
or  \(\log \left|\frac{\sec v}{v x^2}\right|=\log \left|C_1\right|\)
or \(\frac{\sec v}{v x^2}= \pm \mathrm{C}_1\)  ... (3)
Replacing v by \(\frac{y}{x}\)in equation (3), we get
\( \frac{\sec \left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)\left(x^2\right)}=\mathrm{C} \text { where, } \mathrm{C}= \pm \mathrm{C}_1 \\ \)
\(\sec \left(\frac{y}{x}\right)=\mathrm{C} x y\)
which is the general solution of the given differential equation.

Let first function be sin ^{– 1}x and second function be \(\frac{x}{\sqrt{1-x^{2}}}\)
First we find the integral of the second function, i.e., \(\int \frac{x d x}{\sqrt{1-x^{2}}}\)
Put t = 1 – x². Then dt = – 2x dx
\(\text { Therefore, } \quad \int \frac{x d x}{\sqrt{1-x^{2}}}=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^{2}}\)
\(\text { Hence, }\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\left(\sin ^{-1} x\right)\left(-\sqrt{1-x^{2}}\right)-\int \frac{1}{\sqrt{1-x^{2}}}\left(-\sqrt{1-x^{2}}\right) d x \)
\(=-\sqrt{1-x^{2}} \sin ^{-1} x+x+\mathrm{C}=x-\sqrt{1-x^{2}} \sin ^{-1} x+\mathrm{C}\)
Alternatively, this integral can also be worked out by making substitution sin^–1 x = θ and then integrating by parts.

We have: \(\frac { dy }{ dx } +\sqrt { \frac { 1-{ y }^{ 2 } }{ 1-{ x }^{ 2 } } } =0.\)
\(\Rightarrow \) \(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } +\frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } =0\)
| Variables Separable
Integrating, \(\int { \frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } } +\int { \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } } =C\)
\(\Rightarrow \) \({ sin }^{ -1 }\quad y+{ sin }^{ -1 }x=C,\)
Which is the reqd. solution.

The given ellipse is \(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Since (1) is symmetrical about both the axes,

Therefore, area of the ellipse = 4 (Shaded area) = 4 (area OAB)
\(But\ area\ OAB=\overset { 3 }{ \underset { 0 }{ \int { } } } xdy\ [Taking\ horizontal\ strips]\)
\(=\overset { 3 }{ \underset { 0 }{ \int { } } } \frac { 2 }{ 3 } \sqrt { { 9-y }^{ 2 } } dy\)
\([\because \frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow \frac { { x }^{ 2 } }{ 4 } =1-\frac { { y }^{ 2 } }{ 9 } \Rightarrow x=\frac { 2 }{ 3 } \sqrt { 9-{ y }^{ 2 } } (\because x>0)]\)
\(=\frac { 2 }{ 3 } \left[ \frac { y\sqrt { 9-{ y }^{ 2 } } }{ 2 } +\frac { 9 }{ 2 } { sin }^{ -1 }\frac { y }{ 3 } \right] _{ 0 }^{ 3 }\)
\(=\frac { 2 }{ 3 } \left[ \left[ \frac { 3 }{ 2 } (0)+\frac { 9 }{ 2 } { sin }^{ -1 }(1) \right] -[0-0] \right] \)
\(=\frac { 2 }{ 3 } \left[ \frac { 9 }{ 2 } \left( \frac { \pi }{ 2 } \right) \right] =\frac { 3\pi }{ 2 } \)
\(\therefore From(2),\ area\ of\ the\ ellipse=4\left( \frac { 3\pi }{ 2 } \right) =6\pi sq.units\)

We know that the slope of the tangent to a curve is given by \(\frac{d y}{d x} \)
 so,\(\frac { dy }{ dx } =\frac { 2x }{ { y }^{ 2 } } \) ... (1)
Separating the variables, equation (1) can be written as
  \({ y }^{ 2 }dy=2xdx\) ... (2)
Integrating both sides of equation (2), we get
\(\int { { y }^{ 2 }dy=2\int { xdx+c } } \)
\(\Rightarrow \) \(\frac { { y }^{ 3 } }{ 3 } ={ x }^{ 2 }+c\) ...(3)
Substituting x = –2, y = 3 in equation (3), we get C = 5.
Substituting the value of C in equation (3), we get the equation of the required curve as         
\(\frac{y^3}{3}=x^2+5 \ \text { or } \ y=\left(3 x^2+15\right)^{\frac{1}{3}}\)

(i) Using the formula, we express
\(x+2=\mathrm{A} \frac{d}{d x}\left(2 x^{2}+6 x+5\right)+\mathrm{B}=\mathrm{A}(4 x+6)+\mathrm{B}\)
Equating the coefficients of x and the constant terms from both sides, we get
\(4 \mathrm{~A}=1 \text { and } 6 \mathrm{~A}+\mathrm{B}=2 \text { or } \mathrm{A}=\frac{1}{4} \text { and } \mathrm{B}=\frac{1}{2}\)
\(\text {Therefore, }\int \frac{x+2}{2 x^{2}+6 x+5}=\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}\)
\(=\frac{1}{4} \mathrm{I}_{1}+\frac{1}{2} \mathrm{I}_{2} \quad \text { (say) } \)
In I₁, put 2x² + 6x + 5 = t, so that (4x + 6) dx = dt
\(\text {Therefore, } \ \mathrm{I}_{1}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_{1}\)
= log | 2x² + 6x + 5 | + C₁
\(\text {and } I_{2}=\int \frac{d x}{2 x^{2}+6 x+5}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{5}{2}} \)
\(=\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}} \)
\(\text {Put } x+\frac{3}{2}=t, \text { so that } d x=d t, \text { we get }\)
\(\mathrm{I}_{2} =\frac{1}{2} \int \frac{d t}{t^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2 \times \frac{1}{2}} \tan ^{-1} 2 t+\mathrm{C}_{2} \)
\(=\tan ^{-1} 2\left(x+\frac{3}{2}\right)+\mathrm{C}_{2}=\tan ^{-1}(2 x+3)+\mathrm{C}_{2} \)
Using (2) and (3) in (1), we get
\(\int \frac{x+2}{2 x^{2}+6 x+5} d x=\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+C\)
\(\text {where, }\mathrm{C}=\frac{\mathrm{C}_{1}}{4}+\frac{\mathrm{C}_{2}}{2} \)
(ii) This integral is of the form given. Let us express
\(x+3=A \frac{d}{d x}\left(5-4 x-x^{2}\right)+B=A(-4-2 x)+B\)
Equating the coefficients of x and the constant terms from both sides, we get
\(-2 \mathrm{~A}=1 \text { and }-4 \mathrm{~A}+\mathrm{B}=3, \text { i.e., } \mathrm{A}=-\frac{1}{2} \text { and } \mathrm{B}=1\)
\(\text {Therefore, } \int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x=-\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}+\int \frac{d x}{\sqrt{5-4 x-x^{2}}}\)
\(=-\frac{1}{2} \mathrm{I}_{1}+\mathrm{I}_{2}\)
In I₁, put 5 – 4x – x² = t, so that (– 4 – 2x) dx = dt.
\(\text {Therefore, } \quad \mathrm{I}_{1}=\int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}+\mathrm{C}_{1}\)
\(=2 \sqrt{5-4 x-x^{2}}+C_{1}\)
\(\text {Now consider } \quad \mathrm{I}_{2}=\int \frac{d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d x}{\sqrt{9-(x+2)^{2}}}\)
Put x+2=t , so that dx=dt . 
\(\text {Therefore, } \quad \mathrm{I}_{2}=\int \frac{d t}{\sqrt{3^{2}-t^{2}}}=\sin ^{-1} \frac{t}{3}+\mathrm{C}_{2}\)
\(=\sin ^{-1} \frac{x+2}{3}+C_{2}\)
Substituting (2) and (3) in (1) , we obtain  \(\)
\(\int \frac{x+3}{\sqrt{5-4 x-x^{2}}}=-\sqrt{5-4 x-x^{2}}+\sin ^{-1} \frac{x+2}{3}+\mathrm{C}, \text { where } \mathrm{C}=\mathrm{C}_{2}-\frac{\mathrm{C}_{1}}{2}\)

(i) We have x² – 6x + 13 = x ²– 6x + 3² – 3² + 13 = (x – 3)² + 4
\(\text { So, } \ \int \frac{d x}{x^{2}-6 x+13}=\int \frac{1}{(x-3)^{2}+2^{2}} d x\)
Let x – 3 = t. Then dx = dt
\(\int \frac{d x}{x^{2}-6 x+13}=\int \frac{d t}{t^{2}+2^{2}}=\frac{1}{2} \tan ^{-1} \frac{t}{2}+C\)
\(=\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+\mathrm{C}\)
(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,
\(3 x^{2}+13 x-10=3\left(x^{2}+\frac{13 x}{3}-\frac{10}{3}\right)\)
\(=3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right] \text { (completing the square) }\)
\(\text { Thus } \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(\text { Put } x+\frac{13}{6}=t \text { . Then } d x=d t . \)
\(\text { Therefore, } \quad \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d t}{t^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(=\frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C, \text { where } C=C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(\text { (iii) We have } \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\int \frac{d x}{\sqrt{5\left(x^{2}-\frac{2 x}{5}\right)}}\)
\(=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} \text { (completing the square) }\)
\(\text { Put } x-\frac{1}{5}=t . \text { Then } d x=d t \text { . }\)
\(\text { Therefore, } \quad \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}}\)
\(=\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+\mathrm{C}\)
\(=\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+\mathrm{C}\)

Given differential equation is
\(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x \quad[\because x \neq 0]\) 
Above equation can be rewritten as,
\(\left(1+x^{2}\right) d y+(2 x y-\cot x) d x=0\) 
\(\Rightarrow\left(1+x^{2}\right) d y=(\cot x-2 x y) d x\) 
On dividing both sides by \(1+x^{2}\) ,we get
\(d y=\frac{\cot x-2 x y}{1+x^{2}} d x\) 
\(\Rightarrow \frac{d y}{d x}=\frac{\cot x}{1+x^{2}}-\frac{2 x y}{1+x^{2}}\) 
\(\Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{\cot x}{1+x^{2}}\) 
which is a linear differential equation of the form of
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=\frac{2 x}{1+x^{2}} \text { and } Q=\frac{\cot x}{1+x^{2}}\) 
Now,\(\mathrm{IF}=e^{\int P d x}=e^{\frac{1}{1+x^{2}} d x}=e^{\log \left|1+x^{2}\right|}=1+x^{2}\) 
\(\left[\because I_{1}=\int \frac{2 x}{1+x^{2}} d x,\right. \text { put } 1+x^{2}=t \Rightarrow 2 x d x=d t\) 
\(\left.\Rightarrow I_{1}=\int \frac{d t}{t}=\log |t|=\log \left|1+x^{2}\right|\right]\)
and the solution of linear differential equation is given by
\(y \times I F=\int(Q \times I F) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \cot x d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\log |\sin x|+C\) 
\(\Rightarrow y=\frac{\log |\sin x|}{1+x^{2}}+\frac{C}{1+x^{2}}\) 
which is the required solution.

Given, \(\frac{d y}{d x}-(3 \cot x) y=\sin 2 x\) 
which is a linear differential equation of the form
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=-3 \cot x \text { and } Q=\sin 2 x\) 
Now, \(\mathrm{IF}=e^{\int P d x}=e^{-3 \int \cot x d x}=e^{-3 \log (\sin x)}=e^{\log (\sin x)^{-2}}\) 
\(=\frac{1}{\sin ^{3} x}\) 
and the required solution is given by
\(\boldsymbol{y} \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C\) 
\(\Rightarrow y \times \frac{1}{\sin ^{3} x}=\int \frac{1}{\sin ^{3} x} \sin 2 x d x+C\) 
\(\Rightarrow y \times \frac{1}{\sin ^{3} x}=2 \int \frac{\sin x \cos x}{\sin ^{3} x} d x+C\) 
\([\because \sin 2 x=2 \sin x \cos x]\) 
\(\Rightarrow \frac{1}{\sin ^{3} x} \times y=2 \int \frac{\cos x}{\sin ^{2} x} d x+C\) 
\(\Rightarrow \frac{y}{\sin ^{3} x}=-2 \operatorname{cosec} x+C\) 
\(\Rightarrow y=-2\left(\frac{1}{\sin x} \times \sin ^{3} x\right)+C \sin ^{3} x\) 
\(\Rightarrow y=-2 \sin ^{2} x+C \sin ^{3} x\) 
Also, given \(y=2 \text { and } x=\frac{\pi}{2}\), therefore from Eq. (i), we get
\(2=-2 \sin ^{2}\left(\frac{\pi}{2}\right)+C \sin ^{3}\left(\frac{\pi}{2}\right)\) 
\(\Rightarrow 2=-2+C \)
\(\Rightarrow C=4 \) 
On putting the value of C in Eq. (i), we get
\(y=-2 \sin ^{2} x+4 \sin ^{3} x \Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x\) 
which is the required solution.

As shown in the Figure, the line y = 3x + 2 meets x-axis at x\(=\frac{-2}{3}\) and its graph lies below x-axis for \(x \in\left(-1, \frac{-2}{3}\right)\) and above x-axis for \(x \in\left(\frac{-2}{3}, 1\right)\)
The required area = Area of the region ACBA + Area of the region ADEA
\(=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x\)
\(=\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}\)

(i) Given, differential equation is (x² - y²)dx + 2xydy = 0
\(\begin{aligned} \Rightarrow \frac{d y}{d x} & =\frac{-\left(x^2-y^2\right)}{2 x y}=\frac{y^2-x^2}{2 x y} \end{aligned}\)
\(\begin{aligned} =\frac{x^2\left(\frac{y^2}{x^2}-1\right)}{2 x y}=\frac{\left(\frac{y}{x}\right)^2-1}{2\left(\frac{y}{x}\right)} \end{aligned}\)
\(\therefore\)  In RHS, degree of numerator and denominator is same
\(\therefore\)  It is a homogeneous differential equation and can be written as
\(\frac{d y}{d x}=g\left(\frac{y}{x}\right)\)
(ii) Given, differential equation is (x² - y²)dx + 2xy dy = 0
\(\Rightarrow \quad \frac{d y}{d x}=-\frac{\left(x^2-y^2\right)}{2 x y}=\frac{y^2-x^2}{2 x y}\) ...(i)
This is a homogeneous differential equation
On putting y = vx \(\Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}\)
\(\therefore\) From Eq (i), we get
\(\begin{aligned} v+x \cdot \frac{d v}{d x} & =\frac{v^2-1}{2 v} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x \frac{d v}{d x} & =\frac{v^2-1}{2 v}-v \end{aligned}\)
\(\begin{aligned} =\frac{v^2-1-2 v^2}{2 v}=\frac{-v^2-1}{2 v} \end{aligned}\)
\(\Rightarrow \frac{2 v}{v^2+1} d v=\frac{-d x}{x}\)
on integrating both sides, we get
log |v² + 1| = -log x + log c
\(\begin{aligned} & \Rightarrow \quad \log \left|\frac{y^2}{x^2}+1\right|=-\log x+\log c \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \log \left|\frac{y^2+x^2}{x^2} \cdot x\right|=\log c \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{y^2+x^2}{x}=c \end{aligned}\)
\(\Rightarrow\)   y² + x² = cx
which is the required solution.

(i) (c) : We have, \(2 \frac{d^{2} y}{d x^{2}}+3 \sqrt{1-\left(\frac{d y}{d x}\right)^{2}-y}=0\) 
\(\therefore \quad 2 \frac{d^{2} y}{d x^{2}}=-3 \sqrt{1-\left(\frac{d y}{d x}\right)^{2}-y}\) 
Squaring both sides, we get
\(4\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=9\left[1-\left(\frac{d y}{d x}\right)^{2}-y\right]\) 
Here, highest order derivative is \(\frac{d^{2} y}{d x^{2}}\) and its power is 2. So, its degree is 2.
(ii) (d) : We have, \(y \frac{d y}{d x}=\frac{x}{\frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}}\) 
\(\Rightarrow y\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d y}{d x}\right)^{4}=x\) 
\(\Rightarrow\) Here, highest order derivative is \(\frac{d y}{d x}\) is So , its order is 1 and degree is 4.
(iii) (a) : We have,\(y^{\prime \prime \prime}+y^{2}+e^{y^{\prime}}=0\) 
\(\frac{d^{3} y}{d x^{3}}+y^{2}+e^{(d y / d x)}=0\) 
Highest order derivative is \(\frac{d^{3} y}{d x^{3}}\) .So, its order is 3.
Also, the given differential cannot be expressed as a polynomial. So, its degree is not defined.
(iv) (c) : The given differential equation is,
\(\sqrt{a+x} \cdot\left(\frac{d y}{d x}\right)+x=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a+x}}\) 
Clearly, degree = 1
(v) (b) : We have \(y \frac{d y}{d x}=\frac{x}{\frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}}\) 
\(\Rightarrow y\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d y}{d x}\right)^{4}=x\) 
\(\Rightarrow\) Here, highest order derivative is \(\frac{d y}{d x}\) ,So , its order is 1 and degree is 4.
(iii) (a) : We have, y'" +y² + e^y = 0
\(\frac{d^{3} y}{d x^{3}}+y^{2}+e^{(d y / d x)}=0\) 
Highest order derivative is \(\frac{d^{3} y}{d x^{3}}\) So, its order is 3.
Also, the given differential cannot be expressed as a polynomial. So, its degree is not defined
(iv) (c) :The given differential equation is,
\(\sqrt{a+x} \cdot\left(\frac{d y}{d x}\right)+x=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a+x}}\) 
Clearly, degree = 1.
(v) (b) : We have \(\left(1+\left(\frac{d y}{d x} \mid\right)^{3}\right)^{\frac{1}{3}}=7 \frac{d^{2} y}{d x^{2}}\) 
\(\therefore\) Order is 2 and degree is 3. 

(i) (c) : For point of intersection, we have sin x = cos x
\(\Rightarrow \frac{\sin x}{\cos x}=1 \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}\) 
(ii) (a) : \(\int_{0}^{\pi / 4} \sin x d x=[-\cos x]_{0}^{\pi / 4}=-\cos \frac{\pi}{4}+\cos 0\) \(=1-\frac{1}{\sqrt{2}}\) 
(iii) (b) : \(\int_{\pi / 4}^{\pi / 2} \cos x d x=[\sin x]_{\pi / 4}^{\pi / 2}=\sin \frac{\pi}{2}-\sin \frac{\pi}{4}\) \(=1-\frac{1}{\sqrt{2}}\) 
(iv) (c) : \(\int_{0}^{\pi} \sin x d x=[-\cos x]_{0}^{\pi}=[-\cos \pi+\cos 0]=2\) 
(v) (b) : \(\int_{0}^{\pi / 2} \sin x d x=[-\cos x]_{0}^{\pi / 2}=\left[-\cos \frac{\pi}{2}+\cos 0\right]\) = 0+1=1

(a) Both A and R are correct; R is the correct explanation of A

(c) Assertion is correct, Reason is incorrect