The given equation is:
\(y=e^{a x}\left[c_1 \cos b x+c_2 \sin b x\right]\) ... (1)
Differentiating both sides of equation (1) with respect to x, we get
\(\frac { dy }{ dx } ={ e }^{ ax }\left[ -b{ c }_{ 1 }sin\quad bx+b\quad { c }_{ 2 }\quad cos\quad bx \right] \) \(+\left[ { c }_{ 1 }\quad cos\quad bx+{ c }_{ 2 }sin\quad bx \right] { e }^{ ax }.a\)
\(\Rightarrow \frac { dy }{ dx } ={ e }^{ ax }\left[ \left( b{ c }_{ 2 }+a{ c }_{ 1 } \right) cos\quad bx+\left( a{ c }_{ 2 }-b{ c }_{ 1 } \right) sin\quad bx \right] \)    .... (2)
Differentiating both sides of equation (2) with respect to x, we get, \( \frac{d^2 y}{d x^2}= e^{a x}\left[\left(b c_2+a c_1\right)(-b \sin b x)+\left(a c_2-b c_1\right)(b \cos b x)\right] +\left[\left(b c_2+a c_1\right) \cos b x+\left(a c_2-b c_1\right) \sin b x\right] e^{a x} \cdot a \)
\(= e^{a x}\left[\left(a^2 c_2-2 a b c_1-b^2 c_2\right) \sin b x+\left(a^2 c_1+2 a b c_2-b^2 c_1\right) \cos b x\right] \)
Substituting the values of \(\frac{d^2 y}{d x^2}, \frac{d y}{d x}\) and y in the given differential equation, we get
\( = \left.e^{a x}\left[a^2 c_2-2 a b c_1-b^2 c_2\right) \sin b x+\left(a^2 c_1+2 a b c_2-b^2 c_1\right) \cos b x\right] -2 a e^{a x}\left[\left(b c_2+a c_1\right) \cos b x+\left(a c_2-b c_1\right) \sin b x\right] +\left(a^2+b^2\right) e^{a x}\left[c_1 \cos b x+c_2 \sin b x\right] \)
\(=e^{a x}\left[\begin{array}{l} \left(a^2 c_2-2 a b c_1-b^2 c_2-2 a^2 c_2+2 a b c_1+a^2 c_2+b^2 c_2\right) \sin b x \\ +\left(a^2 c_1+2 a b c_2-b^2 c_1-2 a b c_2-2 a^2 c_1+a^2 c_1+b^2 c_1\right) \cos b x \end{array}\right]\)
\(=e^{a x}[0 \times \sin b x+0 \cos b x]=e^{a x} \times 0=0=\text { R.H.S. }\)
Hence, the given function is a solution of the given differential equation.

The given differential equation can be written as
\(\frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}\) ... (1)
It is a differential equation of the form
\(\frac{d y}{d x}=\mathrm{F}(x, y)\) 
\(\mathrm{F}(x, y)=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}\)
Replacing x by \(\lambda x \text { and } y \text { by } \lambda y, \) we get
\(\mathrm{F}(\lambda x, \lambda y)=\frac{\lambda\left[y \cos \left(\frac{y}{x}\right)+x\right]}{\lambda\left(x \cos \frac{y}{x}\right)}=\lambda^0[\mathrm{~F}(x, y)]\)
Thus, F(x, y) is a homogeneous function of degree zero.
Therefore, the given differential equation is a homogeneous differential equation. To solve it we make the substitution
y = vx ... (2)
Differentiating equation (2) with respect to x, we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\) ...(3)
Substituting the value of y and \(\frac{d y}{d x}\) in equation (1), we get
\(\Rightarrow\)\(v+x\frac { dv }{ dx } =\frac { v\quad cosv+1 }{ cos\quad v } \)
or \(x\frac { dv }{ dx } =\frac { 1 }{ cos\quad v } \)
or \(cos\ v dv =\frac { dx }{ x } \) 
Therefore, \(\int \cos v d v=\int \frac{1}{x} d x\)
\(\Rightarrow\) \(sinv =log|x|+log|c|\)
\(\Rightarrow\) \(sinv= log|cx|\)
Replacing v by \(\frac{ y}{ x}\) we get
\(\Rightarrow\) \(sin\left( \frac { y }{ x } \right) = log|cx|,\)
which is the general solution of the differential equation (1).

The given differential equation can be expressed as
\(d y^*=\left(\frac{2 x^2+1}{x}\right) d x^*\)
or \(d y=\left(2 x+\frac{1}{x}\right) d x\).... (1)
\(\Rightarrow \) \(y=\frac { 2{ x }^{ 2 } }{ 2 } +log|x|+c\)
Integrating both sides of equation (1), we get
\(\int d y=\int\left(2 x+\frac{1}{x}\right) d x\)                   
or  \(y=x^2+\log |x|+C\)... (2)
Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we get C = 0.
Now substituting the value of C in equation (2) we get the equation of the required curve as y = x² + log |x|.

We have: \(\frac { dx }{ dy } =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } \) .(1)
Here \(f\left( x,y \right) =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } .\)
\(=\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } ={ \lambda }^{ 0 }f\left( x,y \right) .\)
Thus \(f(x,y)\) is homogeneous function of degree 0.
To solve:
Put  \(\frac { x }{ y } =v\quad i.e\quad x=vy\)
So that   \(\frac { dx }{ dy } =v+y\frac { dy }{ dy } .\)     
\(\therefore\) (1) becomes : \(v+y\frac { dy }{ dx } =\frac { 2vy\quad { e }^{ v }-y }{ 2y\quad { e }^{ v } } \)
\(\Rightarrow \)  \(y\frac { dv }{ dy } =\frac { 2\quad v\quad { e }^{ v }-1 }{ 2\quad { e }^{ v } } -v\)
\(\Rightarrow\)  \(y\frac { dv }{ dy } =\frac { 2\quad v\quad { e }^{ v }-1 }{ 2\quad { e }^{ v } } -v\)
\(\Rightarrow \)  \(y\frac { dv }{ dy } =\frac { -1 }{ 2{ e }^{ v } } \)
\(\Rightarrow \)  \(2{ e }^{ v }dv=-\frac { dy }{ y } \)
Integrating,\(2\int { { e }^{ v }\ dv=-log|y|+c } \)
\(\Rightarrow \) \(2{ e }^{ v }=-log|y|+c\)
\(\Rightarrow \)\(2{ e }^{ \frac { x }{ y } }=-log|y|+c\) ...(2)
When \(x=0,y=1\)
\(\therefore\)  \(2\left( 1 \right) =-log|1|+c\)
\(\Rightarrow \) \(2=-0+c\Rightarrow =2.\)
Putting in (2),
\(2{ e }^{ \frac { x }{ y } }+log|y|=2,\)
Which is the required solution.

(1+x²)y = tan^-1x - \(\frac {\pi}{4}\) is the required solution.

Given differential equation is
\(\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)\) 
which is a homogeneous differential equation as \(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\) .
On putting \(y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\) in equation (i), we get
\(v+x \frac{d v}{d x}=v-\operatorname{cosec} v\) 
\(\Rightarrow x \frac{d v}{d x}=-\operatorname{cosec} v \)
\(\Rightarrow \sin v d v=-\frac{d x}{x}\) 
On integrating both sides, we get
\(\int \sin v d v=-\int \frac{d x}{x}\) 
\(\Rightarrow -\cos v=-\log |x|+C\) 
\(\Rightarrow \cos v=\log |x|-C\) 
\(\Rightarrow \cos \left(\frac{y}{x}\right)=\log |x|-C \quad\left[\text { put } v=\frac{y}{x}\right]..(ii)\) 
Also, given y = 0, when x = 1
Then, \(\cos 0=\log 1-C \Rightarrow 1=0-C \Rightarrow C=-1\) 
So, equation (ii) becomes \(\cos \left(\frac{y}{x}\right)=\log |x|+1\) 
which is the required equation 

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.

Given, \({ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0\)
\(\therefore \ \frac { dy }{ dx } =\frac { -\left( xy+{ y }^{ 2 } \right) }{ { x }^{ 2 } } \)
Put  y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \) The differential equation becomes
\(v+x\frac { dv }{ dx } =-\left( v+{ v }^{ 2 } \right) \)
\(\Rightarrow \frac { dv }{ { v }^{ 2 }+2v } =\frac { dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ \left( v+1 \right) ^{ 2 }-{ 1 }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow \frac { 1 }{ 2 } \log { \frac { v }{ v+2 } } =-\log { x } +\log { C } \)
\(\Rightarrow \frac { C }{ x } =\sqrt { \frac { y }{ y+2x } } \)
If x = 1, y = 1 then \(c=\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } x } =\sqrt { \frac { y }{ y+2x } } \)

y = e^x + 1
Differentiating both sides of this equation with respect to x, we get:
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{x}+1\right) \)
\(\Rightarrow y^{\prime}=e^{x} \)
Now, differentiating equation (1) with respect to x, we get:
\(\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^{x}\right) \)
\(\Rightarrow y^{\prime \prime}=e^{x} \)
Substituting the values of  y' and y'' in the given differential equation, we get the L.H.S. as: 
\(y^{\prime \prime}-y^{\prime}=e^{x}-e^{x}=0=\text { R.H.S. }\)
Thus, the given function is the solution of the corresponding differential equation.

The given differential equation is:
y' + 5y = 0
The highest order derivative present in the differential equation is y'. Therefore, its order is one.
It is a polynomial equation in y'. The highest power raised to y' is 1. Hence, its degree is one.

The given differential equation is:
\(\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0 \)
\(\Rightarrow \frac{\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y}{\tan x \tan y}=0\)
\(\Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\)
\(\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y\)
Integrating both sides of this equation, we get:
\(\int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y \)
\(\therefore \frac{d}{d x}(\tan x)=\frac{d t}{d x} \)
\(\Rightarrow \sec ^{2} x=\frac{d t}{d x} \)
\(\text {Now, } \int \frac{\sec ^{2} x}{\tan x} d x =\int_{t}^{1} d t . \)
\( =\log t \)
\( =\log (\tan x) \)
\(\text {Let } \sec ^{2} x d x=d t \)
\(\text {Similarly, } \int \frac{\sec ^{2} x}{\tan x} d y=\log (\tan y)\)
Substituting these values in equation (1), we get:
\(\log (\tan x)=-\log (\tan y)+\log \mathrm{C} \)
\(\Rightarrow \log (\tan x)=\log \left(\frac{\mathrm{C}}{\tan y}\right) \)
\(\Rightarrow \tan x=\frac{\mathrm{C}}{\tan y} \)
\(\Rightarrow \tan x \tan y=\mathrm{C} \)
This is the required general solution of the given differential equation.\(\)

The given differential equation is:
\(y \log y d x-x d y=0 \)
\(\Rightarrow y \log y d x=x d y \)
\(\Rightarrow \frac{d y}{y \log y}=\frac{d x}{x} \)
Integrating both sides, we get:
\(\int \frac{d y}{y \log y}=\int \frac{d x}{x} \)
\(\text { Let } \log y=t \)
\(\therefore \frac{d}{d y}(\log y)=\frac{d t}{d y} \)
\(\Rightarrow \frac{1}{y}=\frac{d t}{d y} \)
\(\Rightarrow \frac{1}{y} d y=d t \)
Substituting this value in equation (1), we get:
\(\int \frac{d t}{t}=\int \frac{d x}{x} \)
\(\Rightarrow \log t=\log x+\log \mathrm{C} \)
\(\Rightarrow \log (\log y)=\log \mathrm{C} x \)
\(\Rightarrow \log y=\mathrm{Cx} \)
\(\Rightarrow y=e^{\mathrm{Cx}} \)
This is the required general solution of the given differential equation.

The given differential equation is:
\(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x} \)
\(\Rightarrow \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=\tan ^{2} \frac{x}{2} \)
\(\Rightarrow \frac{d y}{d x}=\left(\sec ^{2} \frac{x}{2}-1\right) \)
Separating the variables, we get:
\(d y=\left(\sec ^{2} \frac{x}{2}-1\right) d x\)
Now, integrating both sides of this equation, we get:
\(\int d y=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x=\int \sec ^{2} \frac{x}{2} d x-\int d x \)
\(\Rightarrow y=2 \tan \frac{x}{2}-x+\mathrm{C} \)
This is the required general solution of the given differential equation.\(\)

\(y=\sqrt{a^{2}-x^{2}}\)
Differentiating both sides of this equation with respect to x, we get:
\(\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^{2}-x^{2}}\right) \)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}\left(a^{2}-x^{2}\right) \)
\(=\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x) \)
\(=\frac{-x}{\sqrt{a^{2}-x^{2}}}\)
Substituting the value of \( \frac{d y}{d x}\) in the given differential equation, we get:
\(\text { L.H.S. } =x+y \frac{d y}{d x} =x+\sqrt{a^{2}-x^{2}} \times \frac{-x}{\sqrt{a^{2}-x^{2}}} \)
\(=x-x =0 =\text { R.H.S }\)
Hence, the given function is the solution of the corresponding differential equation. \(\)