Find the solution of the differential equation (d² y/dx²) + 2(dy/dx) + 4y = 2x² + 3e^{-x}: / अवकल समीकरण (d² y/dx²) + 2(dy/dx) + 4y = 2x² + 3e^{-x} का हल ज्ञात कीजिए:

y = e^{-x}[C₁ {3}x + C₂ {3}x] + (1/2)x² - (1/2)x + e^{-x}

y = 4[C₁ {3}x + C₂ {3}x] + (1/2)x² + (1/2)x - e^{-x}

y = e^x[C₁ {3}x + C₂ {3}x] - (1/2)x² + (1/2)x - e^{-x}

Solve the differential equation 3e^x y\, dx + (1 - e^x)² y\, dy...

Q.1

ICAR AIEEA PG 2024 - Statistical Sciences (Q.107)

MCQ (Single Correct)

Solve the differential equation $3e^x \tan y\, dx + (1 - e^x)\sec^2 y\, dy = 0$:

अवकल समीकरण $3e^x \tan y\, dx + (1 - e^x)\sec^2 y\, dy = 0$ का हल ज्ञात कीजिए:

$\frac{\cos y}{(1 - e^x)^3} = c$ $\frac{\cos y}{(1 - e^x)^3} = c$ $\frac{\tan y}{(1 - e^x)^3} = c$ $\frac{\tan y}{(1 - e^x)^3} = c$ $\frac{\sin y}{(1 - e^x)^3} = c$ $\frac{\sin y}{(1 - e^x)^3} = c$ $\frac{\sin y}{(1 - e^x)^5} = c$ $\frac{\sin y}{(1 - e^x)^5} = c$

Q.2

ICAR AIEEA PG 2024 - Statistical Sciences (Q.108)

MCQ (Single Correct)

Find the solution of the differential equation $\frac{dx}{dy} + \frac{x}{y} = y^2$:

अवकल समीकरण $\frac{dx}{dy} + \frac{x}{y} = y^2$ का हल ज्ञात कीजिए:

$x = \frac{1}{4}x^3 + \frac{c}{y}$ $x = \frac{1}{4}x^3 + \frac{c}{y}$ $y = \frac{1}{4}x^3 + \frac{c}{x}$ $y = \frac{1}{4}x^3 + \frac{c}{x}$ $x = \frac{1}{4}y^3 + \frac{c}{x}$ $x = \frac{1}{4}y^3 + \frac{c}{x}$ $x = \frac{1}{4}y^3 + \frac{c}{y}$ $x = \frac{1}{4}y^3 + \frac{c}{y}$

Q.3

ICAR AIEEA PG 2024 - Statistical Sciences (Q.109)

MCQ (Single Correct)

Find the solution of the differential equation $\frac{d^2 y}{dx^2} + 2\frac{dy}{dx} + 4y = 2x^2 + 3e^{-x}$:

अवकल समीकरण $\frac{d^2 y}{dx^2} + 2\frac{dy}{dx} + 4y = 2x^2 + 3e^{-x}$ का हल ज्ञात कीजिए:

$y = e^{-x}[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x] + \frac{1}{2}x^2 - \frac{1}{2}x + e^{-x}$ $y = e^{-x}[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x] + \frac{1}{2}x^2 - \frac{1}{2}x + e^{-x}$ $y = 4[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x] + \frac{1}{2}x^2 + \frac{1}{2}x - e^{-x}$ $y = 4[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x] + \frac{1}{2}x^2 + \frac{1}{2}x - e^{-x}$ $y = e^x[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x] - \frac{1}{2}x^2 + \frac{1}{2}x - e^{-x}$ $y = e^x[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x] - \frac{1}{2}x^2 + \frac{1}{2}x - e^{-x}$ $y = e^x[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x]$ $y = e^x[C_1 \cos \sqrt{3}x + C_2 \sin \sqrt{3}x]$

Q.4

ICAR AIEEA PG 2024 - Statistical Sciences (Q.112)

MCQ (Single Correct)

The solution of the initial value problem $\frac{dy}{dx} = -6xy; y(0) = 7$ is given by:

प्रारंभिक मान समस्या (initial value problem) $\frac{dy}{dx} = -6xy; y(0) = 7$ का हल है:

$7e^{-3x^2}$ $7e^{-3x^2}$ $3e^{-7x^2}$ $3e^{-7x^2}$ $7e^{-x^2}$ $7e^{-x^2}$ $3e^{-x^2}$ $3e^{-x^2}$

Q.5

ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper (Q.101)

MCQ (Single Correct)

If $\theta = t^n e^{-\frac{r^2}{4t}}$, then the value of $n$ that will make $\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = \frac{\partial \theta}{\partial t}$ is given by :

यदि $\theta = t^n e^{-\frac{r^2}{4t}}$ है, तो $n$ का वह मान जो समीकरण $\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = \frac{\partial \theta}{\partial t}$ को संतुष्ट करेगा, होगा :

0 0 1 1 $-\frac{1}{2}$ $-\frac{1}{2}$ $-\frac{3}{2}$ $-\frac{3}{2}$

Given $\theta = t^n e^{-r^2/(4t)}$. Let $u = -r^2/(4t)$, so $\theta = t^n e^u$.
Calculate spatial derivatives:
$\frac{\partial \theta}{\partial r} = t^n e^u \left(-\frac{2r}{4t}\right) = -\frac{r}{2} t^{n-1} e^u$.
Then $r^2 \frac{\partial \theta}{\partial r} = -\frac{1}{2} t^{n-1} r^3 e^u$.
Differentiating with respect to $r$:
$\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = -\frac{1}{2} t^{n-1} \left[3r^2 e^u + r^3 e^u \left(-\frac{r}{2t}\right)\right] = -\frac{1}{2} t^{n-1} e^u \left[3r^2 - \frac{r^4}{2t}\right]$.
Dividing by $r^2$:
$\text{LHS} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = e^u \left[-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
Now calculate time derivative:
$\text{RHS} = \frac{\partial \theta}{\partial t} = n t^{n-1} e^u + t^n e^u \left(\frac{r^2}{4t^2}\right) = e^u \left[n t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
Equating LHS and RHS:
$-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2} = n t^{n-1} + \frac{r^2}{4} t^{n-2}$.
Comparing coefficients of $t^{n-1}$ yields:
$n = -\frac{3}{2}$.

Detailed analysis of options:

0 is incorrect.
is incorrect.
$-1/2$ corresponds to 1D diffusion, not 3D spherical coordinates.
$-3/2$ is the correct value corresponding to 3-dimensional spherical diffusion.

दिया गया है $\theta = t^n e^{-\frac{r^2}{4t}}$।
त्रिज्यीय अवकलज (radial derivatives) ज्ञात करने पर:
$\frac{\partial \theta}{\partial r} = t^n e^{-\frac{r^2}{4t}}\left(-\frac{2r}{4t}\right) = -\frac{r}{2} t^{n-1} e^{-\frac{r^2}{4t}}$.
$r^2 \frac{\partial \theta}{\partial r} = -\frac{1}{2} t^{n-1} r^3 e^{-\frac{r^2}{4t}}$.
$r$ के सापेक्ष पुनः अवकलन करके $r^2$ से भाग देने पर:
$\text{LHS} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = e^{-\frac{r^2}{4t}}\left[-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
अब समय के सापेक्ष अवकलज (time derivative):
$\text{RHS} = \frac{\partial \theta}{\partial t} = n t^{n-1} e^{-\frac{r^2}{4t}} + t^n e^{-\frac{r^2}{4t}}\left(\frac{r^2}{4t^2}\right) = e^{-\frac{r^2}{4t}}\left[n t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
दोनों पक्षों (LHS और RHS) की तुलना करने पर:
$-\frac{3}{2} t^{n-1} = n t^{n-1} \implies n = -\frac{3}{2}$.

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