Register

ISI2017-DCG-24

• retagged by
321 views
0 votes
0 votes

The differential equation $x \frac{dy}{dx} -y=x^3$  with $y(0)=2$  has

  1. unique solution
  2. no solution
  3. infinite number of solutions
  4. none of these
• retagged by
User Avatar

Please log in or register to answer this question.

Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
1 answer
2
go_editor asked Sep 15, 2018
555 views
ISI2017-MMA-9
A function $y(x)$ that satisfies $\dfrac{dy}{dx}+4xy=x$ with the boundary condition $y(0)=0$ is $y(x)=(1-e^x)$ $y(x)=\frac{1}{4}(1-e^{-2x^2})$ $y(x)=\frac{1}{4}(1-e^{2x^2})$ $y(x)=\frac{1}{4}(1-\cos x)$
A function $y(x)$ that satisfies $\dfrac{dy}{dx}+4xy=x$ with the boundary condition $y(0)=0$ is$y(x)=(1-e^x)$$y(x)=\frac{1}{4}(1-e^{-2x^2})$$y(x)=\frac{1}{4}(1-e^{2x^2})$...
User Avatar
0 votes
0 votes
1 answer
3
Sayan Bose asked May 6, 2019
1,125 views
ISI2019-MMA-6
The solution of the differential equation $\frac{dy}{dx} = \frac{2xy}{x^2-y^2}$ is $x^2 + y^2 = cy$, where $c$ is a constant $x^2 + y^2 = cx$, where $c$ is a constant $x^2 – y^2 = cy$ , where $c$ is a constant $x^2 - y^2 = cx$, where $c$ is a constant
The solution of the differential equation $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}$$is$x^2 + y^2 = cy$, where $c$ is a constant$x^2 + y^2 = cx$, where $c$ is a constant$x^2 ...
User Avatar
0 votes
0 votes
1 answer
4
gatecse asked Sep 18, 2019
273 views
ISI2016-DCG-67
The general solution of the differential equation $2y{y}'-x=0$ is (assuming $C$ as an arbitrary constant of integration) $x^{2}-y^{2}=C$ $2x^{2}-y^{2}=C$ $2y^{2}-x^{2}=C$ $x^{2}+y^{2}=C$
The general solution of the differential equation $2y{y}'-x=0$ is (assuming $C$ as an arbitrary constant of integration)$x^{2}-y^{2}=C$$2x^{2}-y^{2}=C$$2y^{2}-x^{2}=C$$x^...
User Avatar
Link Copied!