JEE MAIN Differential Equation Previous Year Questions (PYQs) – Page 14 of 16
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 152 questions
Let y = y(x) be the solution of the differential equation $\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$, with $y\left( {{\pi \over 4}} \right) = 0$. Then, the value of ${(y(0) + 1)^2}$ is equal to :
1
2
3
4
Solution
The solution of the differential equation $\dfrac{dy}{dx}=-\left(\dfrac{x^{2}+3y^{2}}{3x^{2}+y^{2}}\right),\ y(1)=0$ is:
1
2
3
4
Solution
For $x \in \mathbb{R},\ x \ne 0$, if $y(x)$ is a differentiable function such that
$x \int_{1}^{x} y(t)\,dt = (x+1) \int_{1}^{x} t\,y(t)\,dt,$
then $y(x)$ equals: (where $C$ is a constant.)
1
2
3
4
Solution
Given:
\(x\int_1^x y(t)\,dt=(x+1)\int_1^x ty(t)\,dt\)
Let
\(A=\int_1^x y(t)\,dt\)
and
\(B=\int_1^x ty(t)\,dt\)
So,
\(xA=(x+1)B\)
Differentiate both sides with respect to \(x\):
\(A+xy=(x+1)xy+B\)
From \(xA=(x+1)B\), we get
\(A=\frac{x+1}{x}B\)
Put this value in the differentiated equation:
\(\frac{x+1}{x}B+xy=(x+1)xy+B\)
\(\frac{B}{x}=x^2y\)
So,
\(B=x^3y\)
Now differentiate again:
\(\frac{d}{dx}(B)=\frac{d}{dx}(x^3y)\)
\(xy=3x^2y+x^3y'\)
\(x^3y'=xy-3x^2y\)
\(y'=\left(\frac{1}{x^2}-\frac{3}{x}\right)y\)
\(\frac{dy}{y}=\left(\frac{1}{x^2}-\frac{3}{x}\right)dx\)
Integrating both sides:
\(\log y=-\frac{1}{x}-3\log x+\log C\)
So,
\(y=\frac{C}{x^3}e^{-1/x}\)
Therefore,
Answer: (c) \(\frac{C}{x^3}e^{-1/x}\)
\(x\int_1^x y(t)\,dt=(x+1)\int_1^x ty(t)\,dt\)
Let
\(A=\int_1^x y(t)\,dt\)
and
\(B=\int_1^x ty(t)\,dt\)
So,
\(xA=(x+1)B\)
Differentiate both sides with respect to \(x\):
\(A+xy=(x+1)xy+B\)
From \(xA=(x+1)B\), we get
\(A=\frac{x+1}{x}B\)
Put this value in the differentiated equation:
\(\frac{x+1}{x}B+xy=(x+1)xy+B\)
\(\frac{B}{x}=x^2y\)
So,
\(B=x^3y\)
Now differentiate again:
\(\frac{d}{dx}(B)=\frac{d}{dx}(x^3y)\)
\(xy=3x^2y+x^3y'\)
\(x^3y'=xy-3x^2y\)
\(y'=\left(\frac{1}{x^2}-\frac{3}{x}\right)y\)
\(\frac{dy}{y}=\left(\frac{1}{x^2}-\frac{3}{x}\right)dx\)
Integrating both sides:
\(\log y=-\frac{1}{x}-3\log x+\log C\)
So,
\(y=\frac{C}{x^3}e^{-1/x}\)
Therefore,
Answer: (c) \(\frac{C}{x^3}e^{-1/x}\)
The general solution of the differential equation
$(x - y^2),dx + y(5x + y^2),dy = 0$ is:
1
2
3
4
Solution
Let $y=y(x)$ be the solution of the differential equation
$\dfrac{dy}{dx}+y\tan x=2x+x^{2}\tan x,\ x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, such that $y(0)=1$. Then:
1
2
3
4
Solution
Let $y = y(x)$ be the solution curve of the differential equation $(1 + x^2)dy + (y - \tan^{-1}x)dx = 0$, $y(0) = 1$. Then the value of $y(1)$ is:
1
2
3
4
Solution
$\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{\tan^{-1}x}{1 + x^2}$
I.F. $= e^{\tan^{-1}x}$
$y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{1 + x^2} dx$
$y \cdot e^{\tan^{-1}x} = \tan^{-1}x \cdot e^{\tan^{-1}x} - e^{\tan^{-1}x} + c$
$y(0) = 1 \Rightarrow c = 2$
$y(1) = \frac{2}{e^4} + \frac{\pi}{4} - 1$
If $x=f(y)$ is the solution of the differential equation $(1+y^{2})+\big(x-2e^{\tan^{-1}y}\big)\dfrac{dy}{dx}=0,\ y\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ with $f(0)=1$, then $f\!\left(\dfrac{1}{\sqrt{3}}\right)$ is:
1
2
3
4
Solution
Let $\alpha$ be a non-zero real number. Suppose $f:\mathbf{R}\to\mathbf{R}$ is a differentiable function such that
$f(0)=2$ and $\displaystyle \lim_{x\to -\infty} f(x)=1$.
If $f'(x)=\alpha f(x)+3$, for all $x\in\mathbf{R}$, then $f(-\log_{e}2)$ is equal to:
1
2
3
4
Solution
If y = y(x) is the solution of the differential equation ${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$ satisfyingy(0) = 1, then a value of y(loge13) is :
1
2
3
4
Solution
Let $f:[0,1]\to\mathbb{R}$ be such that $f(xy)=f(x)\,f(y)$ for all $x,y\in[0,1]$, and $f(0)\ne 0$.
If $y=v(x)$ satisfies the differential equation $\dfrac{dy}{dx}=f(x)$ with $y(0)=1$, then $y\!\left(\dfrac{1}{4}\right)+y\!\left(\dfrac{3}{4}\right)$ is equal to:
1
2
3
4
Solution
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
