CUET UG Mathematics Applied Differential Equation Previous Year Questions (PYQs)
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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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Solution
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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Solution
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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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,
$2a+6b=2(2)+6(1)$
$=4+6$
$=10$
The solution of the differential equation $xdy-ydx=0$ represent family of
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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.
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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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 $x=1$
$\int_{-1}^{1} e^{|x|},dx=$
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Solution
Since $|x|$ is symmetric,
$\int_{-1}^{1} e^{|x|},dx=2\int_{0}^{1} e^x,dx$
$=2[e^x]_0^1$
$=2(e-1)$
$\frac{d}{dx}\left[\int_{0}^{2a} f(\sin 2x)\,dx\right] =$
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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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CUET UG Mathematics Applied
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