\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)

order – 2 : degree 2

This is a linear differential equation
Here P = 1, Q = e^-x
\(\therefore \int { p\ dx } =\int { 1.dx } =x\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ x }\)
The solution is
\({ ye }^{ \int { pdx } }\int { { Qe }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ -x }.{ e }^{ x }dx+c=\int { dx+c } } \)
\(\int { y{ e }^{ x }=x+c } \)

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.

Here, the highest order derivative is \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } \) whose power is 3.
Therefore, the given differential equation is of order 4 and degree 3.

Given y = e^−x + mx + n .... (1) 
Derentiating cquation (1) wr.t 'x', we get
\(\frac { dx }{ dx } =-e^{ -x }(-1)+m\)
\(\frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-e^{ -x }+m\)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ e^{ x } }+0\)
\(\Rightarrow \left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) =e^x\)
Substituting the value of \(\frac{d^2y}{dx^2}\) in the given differential equation, we get
\(\Rightarrow e^{ x }\left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) -1=e^x(e^x)-1\\= e^{x-x}-1\\ e^0-1= 1-1=0\)
Thus the solution of the given differential equation e^x \(\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \) -1 = 0

The given differential equation is 
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3\times2 }= { 1+\left( \frac { dy }{ dx } \right) } \)
squaring both sides, we get
\({ \left( \frac { dy }{ dx } \right) }^{ 6 }=1+\left( \frac { dy }{ dx } \right) \)
In this equation, the highest order derivative is 2 and its power is 6.
∴ Order 2, degree 6.

\(\frac { dy }{ dx } +xy=cotx\)
Given differential equation is
\(\frac { dy }{ dx } +xy=cotx\)
The highest derivative is 1 and its power is 1 order 1, degree 1.

(i) 65.33^oC
(ii) 53.46

\(y=\frac { 2x-{ x }^{ 2 } }{ 2(1+sinx) } ,x\neq n\pi +(-1)^{ n }\frac { \pi }{ 2 } ,\)∀ n ε Z

\(y=\frac { 2x }{ 1+log|x| } ,x\neq 0\)

\({ tan }^{ -1 }y=\frac { \pi }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 2 } \left( 1+logx \right) ^{ 2 }\)

The given differential cquation may be written as
\(\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0  \)
\(\left(1+x+x y^2\right) \frac{d y}{d x}=-\left(y+y^3\right)  \)
\(\left(1+x+x y^2\right)=-1\left(y^2+1\right) \frac{d x}{d y}  \)
\(y\left(y^2+1\right) \frac{d x}{d y}+1+x\left(y^2+1\right)=0\)
Divided by \( y\left(y^2+1\right) ,\) 
\(\frac{d x}{d y}+\frac{1}{y\left(y^2+1\right)}+\frac{x\left(y^2+1\right)}{y\left(y^2+1\right)}   =0  \)
\(\frac{d x}{d y}+\frac{x}{y}  =-\frac{1}{y\left(y^2+1\right)}\)
This is the form of \( \frac{d x}{d y}+\mathrm{Px}=\mathrm{Q}  \) where \( \mathrm{P}=\frac{1}{y} and  \mathrm{Q}=\frac{-1}{y\left(1+y^2\right)}\)
\(\text { I.F }=e^{\int P d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y\)
So, the solution of the equation is given by
\(x \times \mathrm{I} . \mathrm{F}  =\int(Q \times I . F) d y+c  \)
\(x \times \mathrm{y}  =\int \frac{-1}{y\left(1+y^2\right)} \times y \times d y+c  \)
\(x \mathrm{y}  =\int \frac{-1}{1+y^2} d y+c=-\int \frac{1}{1+y^2} d y+c\)
xy = -tan^-1y + c
xy + tan^-1y = c
Which is the required solution.

Given slope curve
\(\Rightarrow \frac { dy }{ dx } =\frac { y-1 }{ { x }^{ 2 }+x } \) ..... (1)
\(\Rightarrow \frac { dy }{ y-1 } =\frac { dx }{ { x }^{ 2 }+x } \)
\( \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^2+x}=\frac{d x}{x^2+2\left(\frac{x}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
\(ie) \frac{d y}{y-1}=\frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Integrating on both sides, we get
 \( \int \frac{d y}{\log (y-1)}=\int \frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\  \text { ie) } \log (y-1)=\left(\frac{1}{2\left(\frac{1}{2}\right)}\right) \log \left[\frac{\left(x+\frac{1}{2}\right)-\left(\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)+\left(\frac{1}{2}\right)}\right]+\log C \\  \log (y-1)=\log \left(\frac{x}{x+1}\right)+\log C\\   ie)  (y-1)=\frac{C x}{x+1}  \) 
\(\Rightarrow log(y-1)=log\left( \frac { cx }{ x+1 } \right) \)
\(\Rightarrow y-1=\frac { cx }{ x+1 } \)
Since the curve passes through (1, 0) we get,
\(0-1=\frac { c }{ 2 } \Rightarrow c=-2\)
\(\Rightarrow y-1=\frac { -2x }{ x+1 } \)
\(\Rightarrow y=1-\frac { -2x }{ x+1 } \)
\(\Rightarrow y=\frac { x+1-2x }{ x+1 } =\frac { 1-x }{ x+1 } \)
\(\therefore y=\frac { 1-x }{ 1+x } \)

Given equation is m \(\frac{dV}{dt}\) = F- kv  
The given equation can be written as
\(\frac { dv }{ F-kv } =\frac { dt }{ m } \)
Now Integrating, we get
\(\int { \frac { dv }{ F-kv } } =\int { \frac { dt }{ m } } \)
\(\int \frac{d V}{F-k V}=\int \frac{d t}{M}   \frac{\log (F-k V)}{-k}=\frac{t}{M}+C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}-k C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}+\log \mathrm{C} \\ \log [\mathrm{F}-\mathrm{kV}]-\log \mathrm{C}=-\frac{k t}{M} \\ \log \left(\frac{F-k V}{C}\right)=-\frac{k t}{M} \\ \frac{F-k V}{C}=e^{\frac{-k t}{M}} \\ \frac{F-k V}{e^{\frac{-t}{M}}}=\mathrm{C} \Rightarrow \mathrm{C}=e^{\frac{k t}{M}}(F-k V) \\\)
Initial condition:
Given V = 0 when t = 0
\(\mathrm{C}=e^{\frac{k(0)}{M}}[\mathrm{~F}-\mathrm{k}(0)] \\ =\mathrm{e}^0[\mathrm{~F}-0] \\ \mathrm{C}=\mathrm{F} \\ \therefore \mathrm{F}=(F-k V) e^{\frac{k t}{M}}\)

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, Spherical rain drop evaporates at a rate proportionate to the Surface Area
\(\frac { dQ }{ dt } \)\(\infty\) S, where V - volume at anytime t' and S surface area at time 't'
\(\frac { dV }{ dt } =-kS\) .......(1)
Where k is constant of proportionality (k > 0). Negative (-) sign due to decrease in volume on evaporation
\(V = \frac { 4 }{ 3 } 4\pi { r }^{ 2 } =and\ S = 4 \pi r^2\)
\(\frac { dV }{ dt } = \frac{4}{3}\times\pi \times 3r^3 \frac{dr}{dt}= 4 \pi r^2 \frac{dr}{dt}\)
Substituting in (1),
\(4\pi { r }^{ 2 } . \frac{dr}{dt}= -k \times 4\pi r^2\)
\(\frac{dr}{dr}= -k\)
This is equation for rate of change of radius with respect to time 't'.

\(xy=\frac { { y }^{ 4 } }{ 4 } +c\)

\(y=2tan\frac { x }{ 2 } -x+c\)

prove

x\(\frac{dy}{dx}\)+2y = x²
\( \Rightarrow \frac { dy }{ dx } +\frac { 2y }{ x } =x\)
This is a linear differential equation
Here \(p=\frac { 2 }{ x } \)and Q = x
\(\int { pdx } =2\int { \frac { 1 }{ x } } =2logx={ logx }^{ 2 }\)
\(I.F={ e }^{ \int { pdx } }={ e }logx^{ 2 }={ x }^{ 2 }\)
∴ The solution is
\({ ye }^{ \int { dx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { x.{ x }^{ 2 }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { { x }^{ 3 }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\frac { { x }^{ 2 } }{ 4 } +c\)

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

Given \(\frac { dy }{ dx } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } =\frac { y }{ x } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y }{ x } -\frac { { y }^{ 2 } }{ { x }^{ 2 } } ...(1)\)
This is a homogeneous differential equation
put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes,
\(v+x\frac { dv }{ dx } =\frac { vx }{ x } -\frac { { v }^{ 2 }{ x }^{ 2 } }{ { x }^{ 2 } } =v-{ v }^{ 2 }\)
\(x\frac { dv }{ dx } =v-{ v }^{ 2 }-v=-v\)
\(\therefore \frac { dv }{ { v }^{ 2 } } =\frac { -dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ { v }^{ 2 } } =-\int { \frac { dx }{ x } } } \)
\(\Rightarrow \frac { -1 }{ v } =-log\quad x+c\)
\(\Rightarrow \frac { -1 }{ \frac { y }{ x } } =-log\quad x+c\)
\(\Rightarrow \frac { -x }{ y } +logx=c\)

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

Equation of family of parabolas with axis as y axis is given by,
(x-0) = 4a(y-k) .... (1)

Given: Vertex at (0, - 1).
Putting k = -1 in (1), we get
⇒ x² = \(\pm\)4a(y + 1) ....(2)
Differentiating with respect to 'x'  
2x = \(\pm\)4a\(\left( \frac { dy }{ dx } \right) \) ....(3)
⇒ 4a = \(\frac { 2x }{ \frac { dy }{ dx } } \) 
\(\frac{x^2}{2x} = \frac{y+1}{\frac{dt}{dx}}\)
ie) \(x \frac{dy}{dx}-2(y+1) =0\)
This is the required differential equation.

Given the circles centre on r-axis & the circle is passing through the origin.
Let it be (r, 0) & its radius r.

Equation of the circle is
(x - a)² + (y - b)² = r²
(x - r)² + (y - 0)² = r²
⇒ x² - 2xr + r² + y² = r²
⇒ x² - 2xr + y² = 0 ...(1)
defferentiating equation (1) with respect to 'x' we get
⇒ 2x - 2r + 2y \(\frac { dy }{ dx } =0\)
⇒ 2x + 2y \(\frac { dy }{ dx } =2r\)
⇒ x + y \(\frac { dy }{ dx } =r\) ...(2)
Substituting r value in equation (1), we get
x² - 2x \(\left( x+y\frac { dy }{ dx } \right) +{ y }^{ 2 }=0\)
\(\Rightarrow \ { x }^{ 2 }-{ 2x }^{ 2 }-2xy\frac { dy }{ dx } +{ y }^{ 2 }=0\)
\(\Rightarrow \ { -x }^{ 2 }{ -2x }y\left( \frac { dy }{ dx } \right) { +y }^{ 2 }\)
Multiply by '-', we get
\(\Rightarrow \ { x }^{ 2 }{ +2x }y\left( \frac { dy }{ dx } \right) { -y }^{ 2 }\) which is the required differential equation.

General equation of a straight line is 
ax + by + c = 0 .......(1)
where a, b, c \(\in\) R.
Since, the lines are non - vertical,we have b \(\neq\) 0
Dividing b' by equation (1),
\(( \frac{a}{b})x+y+(\frac{c}{b}) = 0 \)
\(Ax+y=C = 0, where A = \frac{a}{b}, C = \frac{c}{b}\) .....(2)
Thus, eventhough 3 arbitrary constants (a, b, c) are present in (1), they can be considered as 2 constants only, as above (2).
Differentiating (1) with respect to x
a + b \(\\ \frac { dy }{ dx } =0\)
Differentiating again with respect to 'x' we get,
(b) \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\quad [\because b\neq 0]\) .....(3)
This is the differential equation of family of all non - vertical lines in a plane.