Explicit form:

$\frac{d y}{d x}=f(x, y)$,

where $x$ is the independent variable and $y$ is dependent on $x$.

$f$ is a given function of two variables.

Implicit form:

$F\left(x, y, \frac{d y}{d x}\right)=0$

where $F$ is a given function of three variables

(1) Variables separable method

Suppose the given ODE can be written in the form $f(x) d x=g(y) d y$.

Then we can simply integrate both sides

$\int f(x) d x=\int g(y) d y$

This gives solutions in implicit form. If possible, we will write $y$ as a function of $x$.

An ODE of the form $\frac{d y}{d x}=f\left(\frac{y}{x}\right)$

In such cases, we put $\frac{y}{x}=u$ ie, $y=u x$

Then $\frac{d y}{d x}=u+x \frac{d u}{d x}$

$\therefore u+x \frac{d u}{d x}=f(u)$

$\Leftrightarrow \quad x \frac{d u}{d x}=f(u)-u$

$\Leftrightarrow \quad \frac{d u}{f(u)-u}=\frac{d x}{x}$. Now integrate both sides

$\frac{d y}{d x}+p(x) y=g(x) \text {, where }$

$p(x)$ and $g(x)$ are given functions of $x$ only. If we multiply the give equation by $e^{\int p(x) d x}$ then the L.H.S. becomes

$e^{\int p(x) d x} \frac{d y}{d x} +p(x) e^{\int p(x) d x} y$

$=\frac{d}{d x} \left[e^{\int p(x) d x} \cdot y\right]$

So, the ODE becomes

$\frac{d}{d x}\left[e^{\int p(x) d x} \cdot y\right]=e^{\int p(x) d x} \cdot g(x)$

$\Rightarrow \quad e^{\int p(x) d x} y=\int e^{\int p(x) d x} g(x) d x+C$

Thus we get the solution.

$e^{\int p(x) d x}$ is called an integrating factor.

If $y=y(r)$ satisfies the differential equation $8 \sqrt{x} \sqrt{9+\sqrt{x}}$

$dy=(\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} d x \quad$ for $x>0$,

and $y(0)=\sqrt{7}$, then find the value of $y(256)$.

Solution- $\quad \int d y=\int \frac{1}{\sqrt{4+\sqrt{9+\sqrt{x}}}} \cdot \frac{1}{8 \sqrt{x} \sqrt{9+\sqrt{x}}} d x$

Put $u=4+\sqrt{9+\sqrt{x}}$

The $d u=\frac{1}{2 \sqrt{9+\sqrt{x}}} \cdot \frac{1}{2 \sqrt{x}} d x$

$y=\int \frac{1}{\sqrt{u}} \cdot \frac{d u}{2}=\sqrt{u}+c$

$s \Rightarrow \quad y=\sqrt{4+\sqrt{9+\sqrt{x}}}+c$

solve $y(0)=\sqrt{7}$, we get $C=0$

$\therefore \quad y(r)=\sqrt{4+\sqrt{9+\sqrt{x}}}$

$\therefore \quad y(256)=\sqrt{4+\sqrt{9+16}}=\sqrt{4+5}=3$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0$.

If $y=f(x)$ satisfies

$\frac{d y}{d x}=(2+5 y)(5 y-2)$, then find the value of $\lim _{x \rightarrow-\infty} f(x)$.

Solution: $\int \frac{1}{(5 y+2)(5 y-2)} d y=\int d x$

$\Rightarrow \int \frac{1}{4}\left(\frac{1}{5 y-2}-\frac{1}{5 y+2}\right) d y=\int d x$

$\Rightarrow \frac{1}{4}\left[\frac{1}{5} \ln |5 y-2|-\frac{1}{5} \ln |5 y+2|\right]=x+c$

$\Rightarrow \quad \ln \left|\frac{5 y-2}{5 y+2}\right|=20 x+20 c$

$y(0)=0 \quad \Rightarrow \quad \ln \left|\frac{0-2}{0+2}\right|=0+20 c$ $\Rightarrow \quad c=0$

$\therefore \quad \ln \left|\frac{5 y-2}{5 y+2}\right|=20 x$

$\Rightarrow \quad\left|\frac{5 y-2}{5 y+2}\right|=e^{20x}$

Suppose $L=\lim _{x \rightarrow-\infty} f(x)$. Then

$\left|\frac{5 L-2}{5 L+2}\right| =\lim _{x \rightarrow-\infty} e^{20 x}=0$ $\Rightarrow \quad 5 L-2 =0 \Rightarrow L=\frac{2}{5},$

$\therefore \lim _{x \rightarrow-\infty} f(x)=\frac{2}{5}$

A curve passes through the point $\left(1, \frac{\pi}{6}\right)$ Let the slope of the tangent to curve at any point $(x, y)$ be

$\frac{y}{x}+\sec \left(\frac{y}{x}\right)$ for $x>0$. Then the equation of the curve is

(A) $\quad \sin \left(\frac{y}{7}\right)=\ln x+\frac{1}{2}$

(B) $\operatorname{cosec}\left(\frac{y}{x}\right)=\ln x+\frac{1}{2}$

(C) $\quad \sec \left(\frac{2 y}{x}\right)=\ln x+2$

(D) $\cos \left(\frac{2 y}{x}\right)=\ln x+\frac{1}{2}$

Solution:${\text{ Give}} \quad \frac{d y}{d x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$

This is a homogenears form

put $y=u x$. Then $u+x \frac{d u}{d x}=u+\sec u$

$\Rightarrow \quad x \frac{d u}{d x}=\sec u$

$\Rightarrow \quad \int \cos u d u=\int \frac{d x}{x}$

$\Rightarrow \quad \sin u=\ln x+c$

$\Rightarrow \quad \sin \left(\frac{y}{x}\right)=\ln x+c$

$\text{when} x=1, y=\frac{\pi}{6} \quad \therefore \quad \sin \frac{\pi}{6}=\ln 1+c$

$\Rightarrow c=\frac{1}{2}$

1) $\sin \left(\frac{y}{x}\right)=\ln x+\frac{1}{2}$

Q4 Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)=2-\frac{f(x)}{x}$ ard $f(1) \neq 1$.

Then

(A) $\lim _{x \rightarrow 0+} f^{\prime}\left(\frac{1}{x}\right)=1$

(B) $\lim _{x \rightarrow 0+} x f\left(\frac{1}{x}\right)=2$

(C) $\lim _{x \rightarrow 0+} x^2 f^{\prime}(x)=0$

(D) $|f(x)| \leq 2$ for all $x \in(0,2)$.

Solution:, write $y=f(x)$

Then $\frac{d y}{d x}=2-\frac{y}{x}$ $\Rightarrow \quad \frac{d y}{d x}+\frac{1}{x} y=2 \leftarrow$ linear ODE.

$\Rightarrow \quad x \frac{d y}{d x}+y=2 x$

$\Rightarrow \quad \frac{d}{d x}(x y)=2 x$

$\Rightarrow \quad x y=x^2+c$

$\Rightarrow \quad y=\frac{x^2+c}{x c}=x+\frac{c}{x}$

Equation $\quad f(x)=x+\frac{c}{x}$ $\Rightarrow f(1)=1+c$

Since $f(1) \neq 1, c \neq 0$.

$\therefore \quad f(x)=x+\frac{c}{x} \text { for some } c \neq 0$

$\Rightarrow f^{\prime}(x)=1-\frac{c}{x^2} \Rightarrow f^{\prime}\left(\frac{1}{x}\right)=1-c x^2$

$\text { i. Option (A) is correct. }$

$x f\left(\frac{1}{x}\right)=x\left[\frac{1}{x}+c x\right]=1+c x^2 \rightarrow 1 \text { as } x \rightarrow 0+$

$\therefore$ (B) is wrong. $x^2 f^{\prime}(x)=x^2\left(1-\frac{c}{\pi^2}\right)=x^2-c$

$\therefore \quad \lim _{x \rightarrow 0} x^2 f^{\prime}(x)=-c \neq 0$

Note that since $f^{\prime}(x)=2-\frac{f(x)}{x}$ $x^2 f^{\prime}(x)=2 x^2-x f(x)$

$\text{Mistake:} \lim _{x \rightarrow 0+} x f(x)=0$

$\lim _{x \rightarrow 0+} f(x)$ need not be finite.

$f(x)=x+\frac{c}{x}, c \neq 0$ Since $\left|\frac{c}{x}\right| \rightarrow+\infty$ as $x \rightarrow 0+$

$\therefore$ $f(x)$ is not bounded on $(0,2)$.

$\therefore$ Optor (D) is wrong.

Solution: $\frac{d y}{d x}+g^{\prime}(x) y=g(x) g^{\prime}(x) \leftarrow \text { linear }$

$\text {Integral factor} =e^{\int g^{\prime}(x) d x}=e^{g(x)}$

$\Rightarrow \quad y \cdot e^{g(x)}=\int e^{g(x)} \cdot g(x) g^{\prime}(x) d x$

By integertion by parts,

$\quad \int g(x)\left(e^{g(x)} g^{\prime}(x) d x\right)=\int g(x) d\left(e^{g(x)}\right)$

$=g(x) e^{g(x)}-\int g^{\prime}(x) \cdot e^{g(x)} d x$

$=g(x) e^{g(x)}-e^{g(x)}+c$

$\therefore \quad y e^{g(x)}=[g(x)-1] e^{g(x)}+c$

$\Rightarrow \quad y=g(x)-1+c e^{-g(x)}$

$y(0)=0 \Rightarrow 0 =g(0)-1+c e^{-g(0)}$

$=0-1+c \Rightarrow c=1$

$\therefore y =g(x)-1+e^{-g(x)}$

$y(2)=g(2)-1+e^{-g(2)}=0-1+e^0=-1+1=0$

$\therefore y(2)=0$

Q6. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a continuous for such that

$f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t \quad$ for all $x \in[0, \infty)$

Then (A) The curve $y=f(x)$ passes throng $(1,2)$

(B) The curve $y=f(x)$ passes through $(2,-1)$

(C) The area of the region $R={(x, y) \in[0,1] \times \mathbb{R}: f(x) \leq y \leq \sqrt{1-x}}$ is $\frac{\pi-2}{4}$.

(D) The area of $R$ is $\frac{\pi-1}{4}$.

Solution: $\quad f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$

$\text {putting x=0, we get f(0)=1}$ $f(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t$

Differentiating with respect to $x$, we get

$f^{\prime}(x) =-2+e^x \int_0^x e^{-t} f(t) d t+e^x \cdot e^{-x} f(x)$

$\Rightarrow f^{\prime}(x) =-2+\int_0^x e^{x-t} f(t) d t+f(x)$ $=-2+[f(x)-1+2 x]+f(x)$

$\Rightarrow \quad f^{\prime}(x)-2 f(x)=2 x-3 \quad \text { Liner ODE.}$

$\text { Integrating factor }=e^{\int-2 d x}=e^{-2 x}$

$\Rightarrow \quad \frac{d}{d x}\left[e^{-2 x} f(x)\right]=e^{-2 x}(2 x-3)$

$\Rightarrow \quad e^{-2 x} f(x)=\int(2 x-3) e^{-2 x} d x$

$=(2 x-3) \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x$

$=-\frac{1}{2}(2 x-3) e^{-2 x}+\frac{e^{-2 x}}{-2}+c$

$\Rightarrow f(x)=-\frac{1}{2}(2 x-3)-\frac{1}{2}+c e^{2 x}$

$\Rightarrow f(x)=c e^{2 x}-x+1$

$f(0)=1 \Rightarrow 1= c+1 \Rightarrow c=0$

$\therefore f(x)=1-x$

$\quad x(1)=1-1=0$

$f(2)=1-2=-1 \quad \therefore \text { (B) is correct }$

$R=\{(x, y) \in[0,1] \times \mathbb{R}: 1-x \leq\sqrt{1-x^2}\}$

Area $(R)=$ ?

$\operatorname{Area}(R) =\frac{1}{4} \pi(1)^2-\frac{1}{2} \times 1 \times 1$

$=\frac{\pi}{4}-\frac{1}{2}=\frac{\pi-2}{4}$

$\therefore$ (C) is corret

(D) is wrong.