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

Given that x² + y² = r², r∈R      ...(1)
The given equation contains exactly one arbitrary constant.
So, we have to differentiate the given equation once. Differentiate (1) with respect to x, we get
2x +2y\(\frac{dy}{dx}\) = 0  which implies \(\frac{dy}{dx}\) = \(-\frac{x}{y}\)
Thus, x² + y² = r² satisfies the differential equation \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).
Hence, x² + y² = r² is a solution of the differential equation  \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \).

The equation of the family of parabolas is given by y² ax = 4, a is an arbitrary constant. ... (1)
Differentiating both sides of (1) with respect to x , we get 2y\(\frac{dy}{dx}=4a\Rightarrow a=\frac{y}{2}\frac{dy}{dx}\) 
Substituting the value of a in (1) and simplifying, we get \(\frac{dy}{dx}=\frac{y}{2x}\) as the required differential equation.

dy + (xy − cos x)dx = 0 is a first order differential equation with degree 1 
since the equation can be rewritten as 
\(\frac{dy}{dx}\) + xy - cos x = 0

Given y = a cos bx  ...(1)
Differentiating equation (1) w.r.t 'x', we get 
\(\frac{d y}{d x}=\mathrm{a}(-\sin \mathrm{b} x) \mathrm{b}=-\mathrm{ab} \sin \mathrm{b} x\)
Again differentiating, we get
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab} \cos \mathrm{b} x \cdot \mathrm{b}   \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab}^2 \cos \mathrm{b} x=-\mathrm{b}^2(\mathrm{a} \cos \mathrm{b} x)  \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{b}^2 \mathrm{y}  \)
\(\frac{d^2 y}{d x^2}+\mathrm{b}^2 \mathrm{y}  =0\)
 Therefore, y = a cos bx is a solution of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ b }^{ 2 }y=0\)

\(\frac { 2log5 }{ lof2 } hours\)

20log_e2years

\( y e^{\left(\frac{1}{r}\right)} \cdot d x   =\left(x e^{\frac{1}{y}}+y\right) d y \)
\(\frac{d x}{d y}   =\frac{x \cdot e^{\left(\frac{6}{y}\right)}+y}{y e^{\left(\frac{x}{y}\right)}} \)
\(\frac{d x}{d y}   =\left(\frac{x}{y}\right)+\frac{1}{e^{\left(\frac{6}{y}\right)}}\)
Put x = \( \mathrm{vy}\) \( \Rightarrow\left(\frac{x}{y}\right)=\mathrm{v}\) and \(\frac{d x}{d y}=\cdot v+y \cdot \frac{d v}{d y} \)
\((1) \Rightarrow \ v+y \cdot \frac{d v}{d y}=v+\frac{1}{e^y} \)
\(\mathrm{e}^v \cdot \mathrm{dv}=\frac{d y}{y}\)
Integrating on both sides,
ie) \(\int e^v \cdot d v   =\int \frac{d y}{y} \)
\(e^v   =\log |y|+\log |c|  \) 
\(e^{\left(\frac{x}{y}\right)}   =\log |c y|\)

\(\\ \\ \\ \frac { dy }{ dx } ={ tan }^{ 2 }(x+y)...(1)\)
Take x + y = t
\(\Rightarrow 1+\frac { dy }{ dx } =\frac { dt }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { dt }{ dx } -1\)
∴ (1) becomes,
\(\frac { dt }{ dx } -1={ tan }^{ 2 }t\)
\(\Rightarrow \frac { dt }{ dx } ={ tan }^{ 2 }t1\)
\(\Rightarrow \frac { dt }{ dx } ={ sec }^{ 2 }(t)\)
\(\Rightarrow \frac { dt }{ { sec }^{ 2 }t } =dx\)
\(\Rightarrow { cos }^{ 2 }t\quad dt=dx\)
\(\left(\frac{1+\cos 2 t^{\circ}}{2}\right) d t=\mathrm{d} x \quad\left(\because \cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\right)\)
\(\left[ cos\quad 2x=2{ cos }^{ 2 }x-1{ cos }^{ 2 }x=\frac { 1+cos2x }{ 2 } \right] \)
Taking integration on both sides, we get
\(\Rightarrow \left( \frac { 1+cos2\quad t }{ 2 } \right) dt=dx\)
\(\Rightarrow \frac { 1 }{ 2 } \int { (1+cos2t)dt=\int { dx } } \)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { sin2t }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { 2sintcost }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } [t+sin\ t\ cost]=x+c\ [\because t=x+y]\)
\(\Rightarrow \frac { 1 }{ 2 } [x+y+sin(x+y)cos(x+y)=x+c\)

\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } .cot\left( \frac { x }{ y } \right) =xdx\)
put \(\frac { x }{ y } =t\)
\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } =dt\)
Substituting these values in equation (1), we get
dt cot(t) = x dx
 cot t dt = ndx
Taking integration on both sides, we get
\(\Rightarrow \int { cot(t)dt=n\int { dx } } \)
\( \int \cot t \mathrm{dt} =n \int d x \)
\(\log (\sin \mathrm{t}) =\mathrm{n} x+\mathrm{C}_1 \)
\(\sin \mathrm{t} =\mathrm{e}^{\mathrm{nx}+\mathrm{c}_1} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{e}^{n x} \mathrm{e}^{\mathrm{C}_r} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{C}^{\mathrm{nx}}\)
\(\\ \Rightarrow sin\left( \frac { x }{ y } \right) ={ e }^{ nx+c }\left[ \because t=\frac { x }{ y } \right] \)

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

y = e^3x (C cos 2x + D sin 2x) .............. (1)
\( \Rightarrow \mathrm{y} \mathrm{}^{-3 x}(-3)+\mathrm{e}^{-3 x} \mathrm{y}^{\prime} \)
\( =-\mathrm{C} 2 \sin 2 x+2 \mathrm{D} \cos 2 x \)
\(\mathrm{e}^{-3 x}\left(-3 \mathrm{y}+\mathrm{y}^{\prime}\right) =-2[\mathrm{C} \sin 2 x-\mathrm{D} \cos 2 x] \)
\(\Rightarrow \mathrm{e}^{-3 x}\left(-3 \mathrm{y}^{\prime}+\mathrm{y}^{\prime \prime}\right) +\left(-3 \mathrm{y}+\mathrm{y}^{\prime}\right)\left(\mathrm{e}^{-3 x} \mathrm{x}-3\right) \)
\( = -2[2 \mathrm{C} \cos 2 x+2 \mathrm{D} \sin 2 x] \)
\( \mathrm{e}^{-3 x}\left[-3 \mathrm{y}^{\prime}+\mathrm{y}^{\prime \prime}+9 \mathrm{y}-3 \mathrm{y}^{\prime}\right]\)
= -4 [C cos 2x + D sin 2x]
\( y^{\prime \prime}-6 y^{\prime}+9 y=-4 e^{3 x}[C \cos 2 x+D \sin 2 x] \)
\( y^{\prime \prime}-6 y^{\prime}+9 y=-4 y \)
\( y^{\prime \prime}-6 y^{\prime}+13 y=0\)

The equation of the given family of curves is
\(y=A e^{-x}+B e^{x}\) .............(1)
where A and B are arbitrary constant
Differentiating equation (1) with respect to 'x', we get
\( \frac{d y}{d x} =\mathrm{Ae}^{-x}(-1)+\mathrm{Be}^{x}(1) \)
\( =-\mathrm{Ae}^{-x}+\mathrm{Be}^{x}\)
Again differentiating equation (2) with respect to 'x', we get
\( \frac{d^{2} y}{d x^{2}}=-A \mathrm{e}^{-x}(-1)+\mathrm{Be}^{x} \)
\( \frac{d^{2} y}{d x^{2}}=A \mathrm{e}^{-x}+\mathrm{Be}^{x 14.3} \)
\( \frac{d^{2} y}{d x^{2}}=\mathrm{y} \) = 0 is a required differential equation

\(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 2 }={ y }^{ 2 }\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 2 }\)

Given \(\frac { dy }{ dx } +y=cosx\)
This is a linear differential equation
Here p = 1, Q = cos x
\(\therefore \int { p\ dx } =\int { dx } =x\)
\(I.F={ e }^{ \int { p\ dx } }={ e }^{ x }\)
The solution is
\({ y }^{ \int { p\ dx } }=\int { Q{ e }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow { ye }^{ x }=\int { cosx.{ e }^{ x }dx+c } \)
\(\Rightarrow { ye }^{ x }=\frac { { e }^{ x } }{ 2 } \left( cosx+sinx \right) +c\)
\(\Rightarrow y=\frac { 1 }{ 2 } \left( cosx+sinx \right) +{ ce }^{ x }\)
\(\therefore \int { { e }^{ ax }cos\ bx\ dx=\frac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \left[ acos\ bx+sin\ ax \right] } \)

Given y = e-2x[A cos 3x- B sin 3x]
⇒ ye²x = A cos 3x-B sin 3x
Differentiating,y¹e²+2y e^2x = -3A sin 3x-3B
cos 3x
Differentiating again we get,
y"e^2x+2(2y')e^2x+4ye^2x = -9(A cos 3x-B sin 3x)
⇒ e^2x( y"+4y'+4y) = -9(A cos 3x - B sin 3x)
⇒ z y"+4y'+4y = -9(A cos 3x-B sin 3x)
⇒ y"+4y'+4y = -9(using (1))
⇒ y"+4y'+13y = 0
is the required differential equation