The given differential equation can be written as
\(\frac{\cos x}{\cos x} \frac{d y}{d x}+y \frac{\sin x}{\cos x}   =\frac{1}{\cos x}  \)
\(\frac{d y}{d x}+\left(\frac{\sin x}{\cos x}\right) y   =\sec x  \)
\(\frac{d y}{d x}+(\tan x) y   =\sec x\)
This is of the form \(\frac{d y}{d x}+P y=Q \)
where
\(\mathrm{P}  =\tan x  \)
\(\mathrm{Q}  =\sec x\)
Thus, the given differential equation is linear.
\(I.F=e^{\int \operatorname{Pdx}}=e^{\int \operatorname{Lin} x d x}=e^{\log (\sec x)}=\sec x\)
So, the required solution is given by 
\({[\mathrm{y} \times \mathrm{I} . \mathrm{F}] }  =\int[Q \times I F] d x+c  \)
\(\mathrm{y} \times \sec x  =\int \sec x \times \sec x d x+c  \)
\(\mathrm{y} \sec x  =\int \sec ^2 x d x+c \\ \mathrm{y} \sec x  =\tan x+\mathrm{c}  \)
\(\div \sec x, \frac{y \sec x}{\sec x}  =\frac{\tan x}{\sec x}+\frac{c}{\sec x} \)
\(\mathrm{y}  =\frac{\sin x}{\cos x} \times \frac{1}{\sec x}+\frac{c}{\sec x}  \)
\(\mathrm{y}  =\frac{\sin x}{\cos x} \times \cos x+c \cos x \)
\(=\sin x+c \cos x\)

The given equation can be written as \(\frac { dx }{ dy } =\frac { \left( \frac { x }{ y } -1 \right) { 2e }^{ x/y } }{ 1+2{ e }^{ x/y } } =g\left( \frac { x }{ y } \right) ..(1)\)
The appearance of \(\frac{x}{y}\) in equation (1), suggests that the appropriate substitution is x = vy.
Put x = vy . Then, we have \(y\frac { dv }{ dy } =-\frac { 2{ e }^{ v }+v }{ 1+2{ e }^{ v } } \)
By separating the variables, we have \(-\frac { 1+2{ e }^{ v } }{ v+2{ e }^{ v } } dv=-\frac { dy }{ y } \)
On integration, we obtain
log |2e^v + v| = −log |y| + log |C| or log |2ye^v+vy| = log |C| or 2ye^v+ vy = ±C.
Replace v by \(\frac{x}{y}\) to get, 2ye^x/y+x = k, where k =土C, Which gives the required solution.

The given differential equation is homogeneous (verify).
Now, we rewrite the given equation in differential form \(\frac { dy }{ dx } =\frac { y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x } \)
Since the initial value of x is 1, we consider x > 0 and take x =\(\sqrt { { x }^{ 2 } } \)
We have \(\frac { dy }{ dx } =\frac { y }{ x } +\sqrt { 1+{ \left( \frac { y }{ x } \right) }^{ 2 } } \)
Let y = vx. Then, \(v+x\frac { dv }{ dx } =v+\sqrt { 1+{ v }^{ 2 } } \), which becomes \(x\frac { dv }{ dx } =\sqrt { 1+{ v }^{ 2 } } \)
By separating variables, we have \(\frac { dv }{ \sqrt { { v }^{ 2 }+1 } } =\frac { dx }{ x } \)
Upon integration, we get \(|v+\sqrt { { v }^{ 2 }+1 } |=log|x|+log|C|\ or\ v+\sqrt { { v }^{ 2 }+1 } =xC\)
Now, we replace v by \(\frac{y}{x}\), we get \(\frac { y }{ x } +\sqrt { \frac { { y }^{ 2 } }{ { x }^{ 2 } } +1 } =Cx\quad (or)\quad y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ Cx }^{ 2 }\) gives the general solution of the given differential equation 
To determine the value of C, we use the condition that y = 0 when x = 1. So, we get C = 1.
Thus \(y=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ x }^{ 2 }\)is the particular solution of the given differential equation.

We know that the given equation is homogeneous
Now, we rewrite the given equation as \(\frac { dy }{ dx } =\frac { 3y }{ 2x } -\frac { x }{ 2y } \)
Taking y = vx , we have \(v+x\frac { dv }{ dx } =\frac { 3v }{ 2 } -\frac { 1 }{ 2v } orx\frac { dv }{ dx } =\frac { { v }^{ 2 }-1 }{ 2v }\)
Separating the variables, we obtain \(\frac { 2vdv }{ { v }^{ 2 }-1 } =\frac { dx }{ x } \)
On integration, we get log\(|{ v }^{ 2 }-1|=log|x|+log|C|,\)
Hence, |v²-1| = |Cx|, where C is an arbitrary constant
Now, replace v by\(\frac{y}{x}\) to get \(|\frac { { y }^{ 2 } }{ { x }^{ 2 } } -1|\) = |Cx|.
Thus, we have |y²-x²| = |Cx³|
Hence, y² − x² = ±Cx³ (or) y² − x2 = kx³ gives the general solution

Given that \(\frac { dy }{ dx } =\frac { x-y+5 }{ 2(x-y)+7 } .\)
Put z = x-y
\(\frac { dz }{ dx } =1-\frac { dy }{ dx } \)
\(\frac { dy }{ dx } =1-\frac { dz }{ dx } \)
Thus, the given equation reduces to
\(1-\frac { dz }{ dx } =\frac { z+5 }{ 2z+7 } \) 
\(\frac { dz }{ dx } =1+\frac { z+5 }{ 2z+7 } \)
\(\frac { dz }{ dx } =\frac { z+2 }{ 2z+7 } \)
Separating the variables, we get
\(\frac { 2z+7 }{ z+2 } dz=dx\)
\(\frac { 2(z+2)+3 }{ z+2 } =dx\)
\(\left( 2+\frac { 3 }{ z+2 } \right) dz=dx\)
Integrating both sides, we get
2z + 3log |z+ 2| = x + C
That is, 2(x − y) + 3log |x −y+2| = x + C

Given that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

A circle passing through the points (a, 0) and (−a, 0) has its centre on y - axis.
Let (0, b) be the centre of the circle. S o, the radius of the circle is \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \) .
Therefore the equation of the family of circles passing through the points (a, 0) and (−a, 0) is x² + ( y − b)² = a² + b², b is an arbitrary constant. ...(1)
Differentiating both sides of (1) with respect to x, we get
\(2x+2(y-b)\frac { dy }{ dx } =0\Rightarrow y-b=\frac { x }{ \frac { dy }{ dx } } \Rightarrow b=\frac { x }{ \frac { dy }{ dx } } +y\)
Substituting the value of b in equation (1), we get
\(2x+2(y-b)\frac { dy }{ dx } =0\Rightarrow y-b=\frac { x }{ \frac { dy }{ dx } } \Rightarrow b=\frac { x }{ \frac { dy }{ dx } } +y\)
\({ x }^{ 2 }+\frac { { x }^{ 2 } }{ { \left( \frac { dy }{ dx } \right) }^{ 2 } } ={ a }^{ 2 }+{ \left[ \frac { x }{ \frac { dy }{ dx } } +y \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ x }^{ 2 }={ a }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left[ x+y{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 2 }\)
\(\Rightarrow ({ x }^{ 2 }-{ y }^{ 2 }-{ a }^{ 2 })\frac { dy }{ dx } -2xy=0\)
which is the required differential equation

The given differential equation can be written as
\(\frac{d y}{d x}+\left(\frac{1}{x}\right) y=\sin x\)
This is of the form \( \frac{d y}{d x}+P y=Q \)
where \(\mathrm{P}=\frac{1}{x} ; \mathrm{Q}=\sin x \)
Thus, the given differential equation is linear.
\( I.F=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x} \dot{=} x \)
So, the required solution is given by
\(\mathrm{y} \times \mathrm{I} . \mathrm{F} =\int(\mathrm{Q} \times I . F) d x+c  \)
\(\mathrm{y} x =\int \sin x \times x d x+c \\ =x(-\cos x)-(-\sin x)+\mathrm{c}\)
yx = - x cos x + sin x
(y + cos x) x = sin x + c is a required solution. 

The given differential equation can be written as
\(\frac { dy }{ dx } +\left( \frac { -x }{ 1-{ x }^{ 2 } } \right) y=\frac { 1 }{ 1-{ x }^{ 2 } } \)
This is of the form \(\frac{dy}{dx}+Py = Q\)
where \( P=\frac { -x }{ 1-{ x }^{ 2 } } ;Q=\frac { 1 }{ 1-{ x }^{ 2 } } \)
Thus, the given differential equation is linear.
\(\int { pdx } =\int { \frac { -x }{ 1-{ x }^{ 2 } } dx } =\frac { 1 }{ 2 } \int { \frac { -2x }{ 1-{ x }^{ 2 } } dx } \)
\(=\frac { 1 }{ 2 } log(1-{ x }^{ 2 })=log\sqrt { 1-{ x }^{ 2 } } \)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log\sqrt { 1-{ x }^{ 2 } } }=\sqrt { 1-{ x }^{ 2 } } \)
So, the required solution is given by 
\({[\mathrm{y} \times \mathrm{I} . \mathrm{F}] }   =\int[Q \times I . F] d x+c \)
\(y \times \sqrt{1-x^2}   =\int \frac{1}{1-x^2} \sqrt{1-x^2} d x+c \)
\( =\int \frac{1}{\sqrt{1-x^2}} \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} d x+c \)
\( =\int \frac{1}{\sqrt{1-x^2}} d x+c \)
\(\div \sqrt{1-x^2}, \frac{y \sqrt{1-x^2}}{\sqrt{1-x^2}}   =\frac{\sin ^{-1} x}{\sqrt{1-x^2}}+\frac{c}{\sqrt{1-x^2}} \)
\(\mathrm{y}   =\frac{\sin ^{-1} x+c}{\sqrt{1-x^2}}+c\left(1-x^2\right)^{-1 / 2}\)
Which is a required solution.

The given equation can be rewritten as \(\frac { dy }{ dx } +\frac { (x\quad tan\quad x-1) }{ x } y=xcosx\)
This is a linear differential equation. Here \(P=\frac { (x\quad tan\quad x-1) }{ x } ;Q=xcosx\)
\(\int { Pdx } =\int { \frac { (xtanx-1) }{ x } } dx=-log|cosx|-log|x|=-log|xcosx|=log\frac { 1 }{ |xcosx| } \)
Thus, \(I.F.={ e }^{ \int { pdx } }={ e }^{ log\frac { 1 }{ |xcosx| } }=\frac { 1 }{ xcosx } \)
Hence the solution is \({ ye }^{ \int { Pdf } }=\int { Q{ e }^{ \int { Pdx } }dx+C } \)
i.e., \(y\frac { 1 }{ xcosx } =\int { (xcosx)\frac { 1 }{ xcosx } dx+C } \)
or \(\\ \\ \\ \\ y\frac { 1 }{ xcosx } =x+C\)
or y = x² cos x + Cx cos x is the required solution.

By putting z = 4x + 2y −1, we have
z' = 4+2y' = 4+2\(\sqrt z\)
hence \(\frac { dz }{ 4+2\sqrt { z } } =dx\).
Integrating, \(\int { \frac { dz }{ 4+2\sqrt { 2 } } =x+C } \)
Putting z = u² , we have
\(\int { \frac { dz }{ 4+2\sqrt { 2 } } =\frac { udu }{ u+2 } =u-2|u+2|+C } \)
or \(\sqrt z\) - 2 In(\(\sqrt z\) + 2) = x+C
from which on substituting z = 4x + 2y −1, we have the general solution
\(\sqrt { 4x+2y-1 } -2\quad In(\sqrt { 4x+2y-1 } +2)=x+C\)

Given that y′ = sin² (x − y +1)
Put z = x-y+1, so that \(\frac { dz }{ dx } =1-\frac { dy }{ dx } \)
Thus, the given equation reduces to 1-\(\frac { dz }{ dx } \) = sin² Z
i.e., \(\frac { dz }{ dx } \) = 1-sin² z = cos²z
Separating the variables leads to \(\frac { dz }{ { cos }^{ 2 }z } =dx\) (or) sec² zdz = dx
On integration, we get tan z = x +C (or) tan (x − y +1) = x +C.

Given that (1 + x³)dy - x² ydx = 0.
The above equation is written as \(\frac { dy }{ y } -\frac { { x }^{ 2 } }{ 1+{ x }^{ 2 } } dx=0\)
Integrating both sides gives log y- \(\frac{1}{3}\)log(1 + x³) = C₁, which implies,
3 log y - log (1 + x³) = log C.
Thus, 3 log y = log (1 + x³) + log C,
which reduces to log y³ = log C(1+x³)
Hence, y³= C (1+x³) gives the general solution of the given differential equation. It is given that when x = 1, y = 2. Then 2³ = C(1 + 1) \(\Rightarrow\) C = 4 and hence the particular solution is y3 = 4(1 + x³).

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution