MAH CET MCA Differential Equation Previous Year Questions (PYQs)
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 4 questions
A company models the rate of change of concentration by
$(y^2+3xy),dx+(x^2+xy),dy=0$, with $y=1$ when $x=1$. Find $y$ as a function of $x$ (choose the correct option).
1
2
3
4
Solution
Put $y=vx$. Then $v+x,dv/dx=-\dfrac{v(v+3)}{1+v}$.
$\displaystyle \frac{1+v}{v(v+2)},dv=-\frac{2,dx}{x} \Rightarrow \ln!\big(v(v+2)\big)=-4\ln x+C$.
$\Rightarrow v(v+2)=\dfrac{C}{x^{4}}$. Using $(x,y)=(1,1)$ gives $C=3$.
$\displaystyle \therefore \frac{y}{x}!\left(\frac{y}{x}+2\right)=\frac{3}{x^{4}} \Rightarrow y=y(x)=x!\left(-1+\sqrt{1+\frac{3}{x^{4}}}\right)$ (positive branch by IC).$\boxed{y=x!\left(-1+\sqrt{1+\dfrac{3}{x^{4}}}\right)}$
A family of parabolas $y=ax^{2}$ $(a>0)$ has vertices on the $x$–axis and axis parallel to $y$–axis. The differential equation of this family is:
1
2
3
4
Solution
$y=ax^{2}$, $a=\dfrac{y}{x^{2}}$. Differentiate: $y'=2ax=\dfrac{2y}{x}$.
$\Rightarrow x,y'-2y=0$.
Given $,\dfrac{dy}{dx}+y=1,$ with $y(0)=2$, find $y(1)$.
1
2
3
4
Solution
Solve linear ODE: $y=Ce^{-x}+1$. Using $y(0)=2\Rightarrow C=1$.
So $y(1)=e^{-1}+1$.
The differential equation of the family of curves $y=e^x(A cosx+B sinx)$, where A and B are arbitrary constants, is
1
2
3
4
Solution
MAH CET MCA
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
MAH CET MCA
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
