New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Differential Equations question

2022 · 25 Jun · Shift 2 · Q32

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Differential Equations
2022 · 25 Jun · Shift 2 · Q32

Differential Equations question

2022 · 25 Jun · Shift 2 · Q32

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

If

y = y(x)

is the solution of the differential equation

2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0

such that

y(e) = {e \over 3}

, then y(1) is equal to :

A
${1 \over 3}$
B
${2 \over 3}$
C
${3 \over 2}$
D
3

View written solutionFree

Correct answer: B

Given differential equation

$2x^2\frac{dy}{dx}-2xy+3y^2=0$

with condition

$y(e)=\frac e3.$

We need to find $y(1)$ .
Rewrite the equation

Divide by $2$ :

$x^2\frac{dy}{dx}-xy+\frac{3}{2}y^2=0$

so

$x^2\frac{dy}{dx}=xy-\frac{3}{2}y^2.$

Hence,

$\frac{dy}{dx}=\frac{y}{x}-\frac{3y^2}{2x^2}.$
Use the substitution

Since the equation is homogeneous in $y$ and $x$ , put

$y=vx$

where $v=v(x)$ .

Then

$\frac{dy}{dx}=v+x\frac{dv}{dx}.$

Substitute into the differential equation:

$v+x\frac{dv}{dx}=v-\frac{3}{2}v^2.$

Therefore,

$x\frac{dv}{dx}=-\frac{3}{2}v^2.$
Separate variables

$\frac{dv}{v^2}=-\frac{3}{2}\frac{dx}{x}.$

Integrate both sides:

$\int v^{-2}\,dv=-\frac{3}{2}\int \frac{dx}{x}.$

$-\frac{1}{v}=-\frac{3}{2}\ln x+C.$

Rearranging,

$\frac{1}{v}=\frac{3}{2}\ln x+C_1.$
Back-substitute $v=\dfrac{y}{x}$

Since

$v=\frac{y}{x},$

we get

$\frac{x}{y}=\frac{3}{2}\ln x+C_1.$
Use the initial condition

Given

$y(e)=\frac e3,$

so

$\frac{e}{e/3}=3.$

Also, $\ln e=1$ . Thus,

$3=\frac{3}{2}(1)+C_1$

$C_1=3-\frac{3}{2}=\frac{3}{2}.$

Hence the solution is
$=\frac{3}{2}(\ln x+1).$$$
Find $y(1)$

At $x=1$ , we have $\ln 1=0$ . So

$\frac{1}{y(1)}=\frac{3}{2}.$

Therefore,

$y(1)=\frac{2}{3}.$
Check options
- A: $\frac{1}{3}$
- B: $\frac{2}{3}$
- C: $\frac{3}{2}$
- D: $3$
The correct option is B.

More from Differential Equations

If $\frac{d y}{d x} + 2 y tan x = sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is :2022 · MCQ
Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ , $x \in (- 1, 1)$ pass through the origin. Then $- \frac{3}{2} \int \frac{3}{2} f (x) d x$ …2022 · MCQ
Suppose $y = y (x)$ be the solution curve to the differential equation $\frac{d y}{d x} - y = 2 - e^{- x}$ such that $x \to \infty lim y (x)$ is finite. If $a$ and $b$ are respectively the $x$ - and $y$ -intercepts of the tangent to…2022 · Numerical
Let the solution curve y = y(x) of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then y(2) is equal to .2022 · Numerical
Let $S = (0, 2 π) - {\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}}$ . Let $y = y (x)$ , x $\in$ S, be the solution curve of the differential equation $\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}, y (\frac{π}{4}) = \frac{1}{2}$ …2022 · Numerical
If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x e^{x}$ , $y (1) = 0$ then the local maximum value of the function $z (x) = x^{2} y (x) - e^{x}, x \in R$ is :2022 · MCQ
If the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of y(2) is…2022 · MCQ
Let $y = y_{1} (x)$ and $y = y_{2} (x)$ be two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0) = 0$ and $y_{2} (0) = 1$ respectively. Then, the number of points of intersection of $y = y_{1} (x)$ and $y = y_{2} (x)$ is2022 · MCQ