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MAH CET MCA Previous Year Questions (PYQs)

MAH CET MCA Differential Equation PYQ

MAH CET MCA PYQ
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).





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ

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).$\boxed{y=x!\left(-1+\sqrt{1+\dfrac{3}{x^{4}}}\right)}$
MAH CET MCA PYQ
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:





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ

Solution

$y=ax^{2}$, $a=\dfrac{y}{x^{2}}$. Differentiate: $y'=2ax=\dfrac{2y}{x}$. $\Rightarrow x,y'-2y=0$.
MAH CET MCA PYQ
Given $,\dfrac{dy}{dx}+y=1,$ with $y(0)=2$, find $y(1)$.





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ

Solution

Solve linear ODE: $y=Ce^{-x}+1$. Using $y(0)=2\Rightarrow C=1$. So $y(1)=e^{-1}+1$.
MAH CET MCA PYQ
The differential equation of the family of curves $y=e^x(A cos⁡x+B sin⁡x)$, where A and B are arbitrary constants, is





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2024 PYQ

Solution



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