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CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Differential Equation PYQ

CUET UG Mathematics Applied PYQ
The degree of the differential equation $\left(1-\left(\dfrac{dy}{dx}\right)^2\right)^{3/2}=k\dfrac{d^2y}{dx^2}$ is:





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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2024 Mathematics PYQ

Solution

CUET UG Mathematics Applied PYQ

Match List-I with List-II :

List-I List-II
(A) Integrating factor of $x\,dy-(y+2x^{2})\,dx=0$ (I) $\dfrac{1}{x}$
(B) Integrating factor of $(2x^{2}-3y)\,dx=x\,dy$ (II) $x$
(C) Integrating factor of $(2y+3x^{2})\,dx+x\,dy=0$ (III) $x^{2}$
(D) Integrating factor of $2x\,dy+(3x^{3}+2y)\,dx=0$ (IV) $x^{3}$

Choose the correct answer from the options given below:






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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2024 Mathematics PYQ

Solution

CUET UG Mathematics Applied PYQ
If $a$ and $b$ are order and degree of differential equation $y''+(y')^2+2y=0$, then value of $2a+6b$ is:





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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Given differential equation:

$y''+(y')^2+2y=0$

Highest order derivative is $y''$

So,
Order $a=2$

Power of highest order derivative $y''$ is 1

So,
Degree $b=1$

Now,

$2a+6b=2(2)+6(1)$
$=4+6$
$=10$
CUET UG Mathematics Applied PYQ
The solution of the differential equation $xdy-ydx=0$ represent family of





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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

$xdy-ydx=0$

Divide by $dx$:

$x\frac{dy}{dx}-y=0$

So,

$x\frac{dy}{dx}=y$

$\frac{dy}{dx}=\frac{y}{x}$

Separate variables:

$\frac{dy}{y}=\frac{dx}{x}$

Integrate:

$\log|y|=\log|x|+C$

$\log\left|\frac{y}{x}\right|=C$

$\frac{y}{x}=k$

$y=kx$

This represents straight lines passing through origin.
CUET UG Mathematics Applied PYQ
For differential equation $y e^{\frac{x}{y}},dx=(x e^{\frac{x}{y}}+y^2),dy$, $y(0)=1$, the value of $x(e)$ is equal to:





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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Given

$y e^{\frac{x}{y}},dx=(x e^{\frac{x}{y}}+y^2),dy$

Divide by $dy$:

$y e^{\frac{x}{y}}\frac{dx}{dy}=x e^{\frac{x}{y}}+y^2$

Notice that

$\frac{d}{dy}\left(x e^{\frac{x}{y}}\right)=y e^{\frac{x}{y}}\frac{dx}{dy}-x e^{\frac{x}{y}}$

So equation becomes

$\frac{d}{dy}\left(x e^{\frac{x}{y}}\right)=y^2$

Integrate:

$x e^{\frac{x}{y}}=\frac{y^3}{3}+C$

Now use initial condition $y(0)=1$

So at $x=0$, $y=1$

$0\cdot e^{0}= \frac{1}{3}+C$

$C=-\frac{1}{3}$

Thus,

$x e^{\frac{x}{y}}=\frac{y^3-1}{3}$

Now put $y=e$

$x e^{\frac{x}{e}}=\frac{e^3-1}{3}$

It satisfies $x=1$
CUET UG Mathematics Applied PYQ
$\int_{-1}^{1} e^{|x|},dx=$





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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Since $|x|$ is symmetric,

$\int_{-1}^{1} e^{|x|},dx=2\int_{0}^{1} e^x,dx$

$=2[e^x]_0^1$

$=2(e-1)$
CUET UG Mathematics Applied PYQ
$\frac{d}{dx}\left[\int_{0}^{2a} f(\sin 2x)\,dx\right] =$





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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Given solution:

$y = e^x (a\cos x + b\sin x)$

Here the solution contains **two arbitrary constants** $a$ and $b$.

The order of a differential equation is equal to the number of arbitrary constants in its general solution.

Therefore,
Observe that the integral

$\int_{0}^{2a} f(\sin 2x)\,dx$

does **not depend on $x$** because the limits are constants.

Hence the entire expression is a constant.

Derivative of a constant is zero.

Therefore,

$\frac{d}{dx}\left[\int_{0}^{2a} f(\sin 2x)\,dx\right] = 0$

Order $=2$


CUET UG Mathematics Applied

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