y = Acos x + Bsin x ... (1)
Differentiating (1) twice successively, we get
\(\frac{dy}{dx}\)= −Asin x + Bcos x. ...(2)
\(\frac{d^2y}{dx^2}\) = -Acos x − Bsin x = −(A cos x + B sin x). ...(3)
Substituting (1) in (3), we get \(\frac{d^2y}{dx^2}\) + = 0 as the required differential equation

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

The equation of the family of all ellipses having foci on the x -axis and centre at the origin is given by \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1,a>b\).
where a and b are arbitrary constants.
Differentiating equation (1) with respect to x, we get
\(\frac { 2x }{ { a }^{ 2 } } +\frac { 2y }{ { b }^{ 2 } } \frac { dy }{ dx } =0\quad \Rightarrow \frac { x }{ { a }^{ 2 } } +\frac { y }{ { b }^{ 2 } } \frac { dy }{ dx } =0\)
Differentiating equation (2) with respect to x, we get 
\(\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] =0\Rightarrow \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] \)
Substituting the value of \(\frac{1}{a^2}\) in equation (2) and simplifying, we get \(-\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] x+\frac { v }{ { b }^{ 2 } } \frac { dy }{ dx } =0\Rightarrow xy\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 }-y\frac { dy }{ dx } =0\) which is the required differential equation

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 equation can be written as \(\frac { dy }{ dx } -\frac{2}{y}x=y^2e^{-y}\).
This is a linear differential equation. Here \(P=-\frac { 2 }{ y } ;Q={ y }^{ 2 }{ e }^{ -y }\)
\(\int { pdy } =\int { -\frac { 2 }{ y } dy=-2log|y|=log{ |y| }^{ -2 } } =log\left( \frac { 1 }{ { y }^{ 2 } } \right) ,\)
Thus, \(I.F={ e }^{ \int { Pdy } }={ e }^{ log\left( \frac { 1 }{ { y }^{ 2 } } \right) }=\frac { 1 }{ { y }^{ 2 } } .\)
Hence the solution is \(x{ e }^{ \int { Pdy } }=\int { Q{ e }^{ \int { Pdy } }dy+C } \)
Thus, \(x\left( \frac { 1 }{ { y }^{ 2 } } \right) =\int { { y }^{ 2 }{ e }^{ -y } } \left( \frac { 1 }{ { y }^{ 2 } } \right) dy+C=\int { { e }^{ -y }dy+C } =-{ e }^{ -y }+C\)or x = −y²e^−y +Cy² is the required solution

Here, to make the coefficient of \(\frac { dy }{ dx } \) unity, divide both sides by (1+x³).
Then the equation is \(\frac { dy }{ dx } +\frac { { 6x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation in y.
Here, \(P=\frac { { 6x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } \)
\(\int { Pdx } =\int { \frac { { x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =2log|1+{ x }^{ 3 }|=log{ |1+{ x }^{ 3 }| }^{ 2 }=log{ ({ 1+x }^{ 3 }) }^{ 2 }\)
Thus, I.F.\(={ e }^{ \int { Pdx } }={ e }^{ log{ ({ 1+x }^{ 3 }) }^{ 2 } }={ (1+{ x }^{ 3 }) }^{ 2 }\)
Hence the solution is \(ye^{ \int { Pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C. } \)
Thus is, \(y{ (1+{ x }^{ 2 }) }^{ 2 }=\int { \frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } } { (1+{ x }^{ 3 }) }^{ 2 }dx+C=\int { (1+{ x }^{ 2 }) } (1+{ x }^{ 3 })dx+C=\int { (1+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 5 }) } dx+C\) \(or\ y{ (1+{ x }^{ 3 }) }^{ 2 }=x+\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +\frac { x^{ 6 } }{ 6 } +C\)
\(and\quad y=\frac { 1 }{ { (1+{ x }^{ 3 }) }^{ 2 } } \left[ x+\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +\frac { x^{ 6 } }{ 6 } +C \right] \)is the 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).