AMU MCA Differential Equation Previous Year Question Paper and Solution with PDF

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AMU MCA Previous Year Questions (PYQs)

AMU MCA Differential Equation PYQ

Qus : 1 AMU MCA PYQ

The solution of the linear difference equation $y_{k+1}-ay_k=0$, $a\ne 1$ is

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

$y_{k+1}=ay_k \Rightarrow y_k=ca^k$

Qus : 2 AMU MCA PYQ

The general solution of the differential equation $(D^2 - a^2)^3 y = e^{ax}$, where $D=\frac{d}{dx}$ is

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Solution

Characteristic equation:

$(m^2-a^2)^3=0$

Roots: $m=\pm a$ each of multiplicity 3.

Hence complementary function:

$(c_1+c_2x+c_3x^2)e^{ax}+(c_4+c_5x+c_6x^2)e^{-ax}$

Since RHS is $e^{ax}$ and $m=a$ has multiplicity 3,

Particular integral = $x^3\frac{e^{ax}}{3!\,(2a)^3} =\frac{x^2e^{ax}}{8a^2}$

Qus : 3 AMU MCA PYQ

Solution of differential equation $y^2\log y = xy\frac{dy}{dx}+\left(\frac{dy}{dx}\right)^2$ is

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Solution

Qus : 4 AMU MCA PYQ

The Laplace transform of $\sin \sqrt{x}$ is:

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Solution

The feasible region allows $z$ to increase indefinitely, hence the objective function is unbounded.

Qus : 5 AMU MCA PYQ

The solution of the differential equation $y\sin 2x,dx - (y^2 + \cos^2 x),dy = 0$ is:

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Solution

After making the equation exact and integrating, the solution obtained is $3y\sin 2x + y^2 + 2y^3 = C$.

Qus : 6 AMU MCA PYQ

The solution of $3\frac{\partial^2 z}{\partial x,\partial y} - 2\frac{\partial^2 z}{\partial y^2} - \frac{\partial z}{\partial y} = 0$ is:

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

Solving the auxiliary equation gives complementary function of the form $\phi_1(x) + e^{x/3}\phi_2(3y+2x)$.

Qus : 7 AMU MCA PYQ

An integrating factor of $\frac{dy}{dx} = \frac{1}{3x + y^2 + 2}$ is:

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

The equation is linear in $x$; integrating factor is $e^{-\int 3,dy} = e^{-3y}$.

Qus : 8 AMU MCA PYQ

The solution of $y' = 2px + \tan^{-1}(xp^2)$ (where $p=\dfrac{dy}{dx}$) is

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Solution

Assume $p=c$ Then $y=cx+\tan^{-1}(c^2x)$

Qus : 9 AMU MCA PYQ

The extremal of functional $\iint_D\left[\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+\left(\frac{\partial^2 z}{\partial x\partial y}\right)^2\right]dxdy$ is

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Solution

Qus : 10 AMU MCA PYQ

The complete solution of $(p^2+q^2)=qz$ is

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Solution

$z=(a^2+b^2)x+by$

Qus : 11 AMU MCA PYQ

The solution of the differential equation $ \dfrac{d^2y}{dx^2} - 2\dfrac{dy}{dx} + y = xe^x \sin x $

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Solution

Auxiliary equation: $ m^2 - 2m + 1 = 0 $ $ (m-1)^2 = 0 $ So CF: $ y_c = (c_1 + c_2 x)e^x $ Particular integral gives: $ y_p = e^x(2\cos x + x\sin x) $ Therefore, $ y = (c_1 + c_2 x)e^x + e^x(2\cos x + x\sin x) $

Qus : 12 AMU MCA PYQ

An integrating factor for $ (y^2+2y),dx+(2xy+x^2),dy=0 $ is

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Solution

$ x^{-1/3}y^{-5/3} $

Qus : 13 AMU MCA PYQ

If $y_1$ and $y_2$ are two solutions of $ y''+p(x)y'+q(x)y=0 $ and Wronskian $W(y_1,y_2)=0$, then $y_1,y_2$ are

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Solution

If Wronskian = 0 ⇒ solutions are linearly dependent.

Qus : 14 AMU MCA PYQ

The general solution of $(y - z)p + (z - x)q = x - y$ is

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Solution

Given Lagrange’s linear PDE $(y - z)p + (z - x)q = x - y$ Auxiliary equations: $\frac{dx}{y - z} = \frac{dy}{z - x} = \frac{dz}{x - y}$ First integral: Add numerators and denominators: $(y - z) + (z - x) + (x - y) = 0$ ⇒ $dx + dy + dz = 0$ ⇒ $x + y + z = c_1$ Second integral: Multiply by $x, y, z$ respectively and add: $x(y - z) + y(z - x) + z(x - y) = 0$ ⇒ $d(x^2 + y^2 + z^2) = 0$ ⇒ $x^2 + y^2 + z^2 = c_2$ Hence general solution is $\phi(x + y + z,; x^2 + y^2 + z^2) = 0$

Qus : 15 AMU MCA PYQ

The solution of $\frac{dx}{x^2 - yz - z^2} = \frac{dy}{2xy} = \frac{dz}{2xz}$ is given by

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Solution

From second and third: $\frac{dy}{2xy} = \frac{dz}{2xz}$ ⇒ $\frac{dy}{y} = \frac{dz}{z}$ ⇒ $\frac{y}{z} = c_1$ Using substitution in first part gives second integral $\frac{x^2 + y^2 - z^2}{2} = c_2$

Qus : 16 AMU MCA PYQ

An integrating factor for $(\cos y \sin 2x)dx + (\cos^2 y - \cos^2 x)dy = 0$

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Using standard form and checking $\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}$ Integrating factor becomes $\frac{1}{\tan^2 y + \sec y \tan y}$

Qus : 17 AMU MCA PYQ

The general solution of $y = 2x \frac{dy}{dx} + y\left(\frac{dy}{dx}\right)^2$ is

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Put $p = \frac{dy}{dx}$

$y = 2xp + yp^2$

$y(1 - p^2) = 2xp$

$y = \frac{2xp}{1 - p^2}$

This is Clairaut’s form. Hence general solution is

$2xc - y + c^2 = 0$

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