The slope of the tangent to the curve at any point = \(\frac { 1 }{ 4(odinate\ at\ the\ point) } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { 1 }{ 4y } \)
The equation can be written as
\(\Rightarrow 4ydy\ =\ dx\) ....(1)
Integrating equation (1) on both sides, we get
\(4\int { y\quad dy } =\int { dx } \)
\(\Rightarrow 4.\frac { { y }^{ 2 } }{ 2 } =x+c\)
\(\Rightarrow { 2y }^{ 2 }=x+c\) ...(2)
Since the curve passes through (2, 5), we get
2(5)² = 2 + c
⇒ 50 - 2 = c
⇒ c = 48
Substituting the value of cincquation (2), we get
 2y² = x + 48 which is the requaired equation of the curve.

given equation of family of curves is
y = Ae^8x + Be^-8x ..(1)
where A & B are arbitrary constants. Differentiating cquation (1) twice successively (because we have two arbitrary constant), we get
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } \\ \\ \) = 8Ae^8x + 8Be^-8x
Differentiating again with respect to 'x' we get,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64Ae^8x + 64Be^-8x
= 64(Ae^8x + Be^-8x)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64 y [using (1)]
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) - 64 y = 0
Which 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.

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

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

Let A be the mass of the isotope remaining after t years, and let −k be the constant of proportionality, where k > 0. Then the rate of decomposition is modeled by \(\frac{da}{dt}=-kA,\) where the minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables,we get\(\frac{da}{dt}=-kdt\).
Integrating on both sides, we get log |A| = −kt + log |C| or A = Ce−^kt.
Given that the initial mass is 200mg. That is, A = 200 when t = 0 and thus, C = 200.
Thus, we get A = − 200e^-kt.
Also, A =150when t = 2 and therefore, k = \(\frac{1}{2}log(\frac{4}{3})\)
Hence, A(t) = 200e^{\(\frac{1}{2}log(\frac{4}{3})\)} is the mass of isotope remaining after t years.
The half-life t_h is the time corresponding to A = 100 mg
Thus, \({ t }_{ k }=\frac { 2log\left( \frac { 1 }{ 2 } \right) }{ log\left( \frac { 3 }{ 4 } \right) } \).

Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 1 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 1 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 1 }{ x } } dx=logx\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ log\quad x }=x\)
\(\therefore\) The solution is \({ e }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }=dx+c } \)
\(u=cos\quad x;dv=x\)
\(du=\frac { 1 }{ x } ,v=\frac { { x }^{ 2 } }{ x } \)
\(\int { udv } =uv-\int { vdu } \)
\(yx=\int { xlogxdx+c } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\int { \frac { { x }^{ 2 } }{ 2 } } .\frac { 1 }{ x } dx\)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } \int { xdx } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } .\frac { { x }^{ 2 } }{ 2 } +c\)
\(\Rightarrow xy=\frac { { 2x }^{ 2 }logx-{ x }^{ 2 }+4c }{ 4 } \)
\(\Rightarrow 4xy=2{ x }^{ 2 }logx-{ x }^{ 2 }+4c\)

⇒ tan y\(\frac{dy}{dx}\) = cos (x+y) + cos (x-y)
⇒ tan y\(\frac{dy}{dx}\) = cos x cos y- sin x sin y + cos x cos y + sin x siny
[∵ cos (A+B) = cos A cos B- sin A sin B cos (A- B) = cos Acos B + sin A sin B]
= 2 cosx cosy
\(\Rightarrow \frac { tan\ y }{ cos\ y } dy=2\ cos\ x\ dx\)
Taking integration on both sides, we get
\(\Rightarrow \int { tan\ y\ sec\ y\ dy=2\int { cos\ x\ dx } } \)
\(\\ \\ \Rightarrow \ sec\ y=2sin\ x+c\)

ydx + (1 + x²) tan^-1 xdy = 0
\(\mathrm{yd} x=-\left(1+x^2\right) \tan ^{-1} x \mathrm{dy} \)
\(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}=-\frac{d y}{y} \)
Take \(\mathrm{t}=\tan ^{-1} x \)
\(\mathrm{dt}=\frac{1}{1+x^2} d x\)
The equation can be written as
\(\frac{d t}{t}=-\frac{d y}{y}\)
Taking Integration on both sides, we get
\(\int \frac{d t}{t}=-\int \frac{d y}{y}\)
log t = - log y + log C
log (tan^-1 x) = -log y+ log C
log (tan^-1 x) + log y = log C
log y(tan^-1 x) = log c
y tan^-1 x = c

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
\(\sqrt { 4x+2y-1 } -2\quad In(\sqrt { 4x+2y-1 } +2)=x+C\)