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}

गणित एवं संख्यात्मक विश्लेषण Practice Questions (Q126-130) - Statistical Sciences

126

ICAR PG Statistical Sciences 29 Jun 2024Shift 1

MCQ (Single Correct) +4 −1

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:

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

127

ICAR PG Statistical Sciences 29 Jun 2024Shift 1

MCQ (Single Correct) +4 −1

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:

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

128

ICAR PG Statistical Sciences 29 Jun 2024Shift 1

MCQ (Single Correct) +4 −1

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:

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

129

ICAR PG Statistical Sciences 09 Jul 2023Shift 1

MCQ (Single Correct) +4 −1

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:
$-\frac{3}{2} t^{n-1} = n t^{n-1} \implies n = -\frac{3}{2}$.

विकल्पों का विस्तृत विश्लेषण:

0 गलत है।
गलत है।
$-1/2$ एक-विमीय विसरण (1D diffusion) के लिए होता है।
$-3/2$ सही उत्तर है (त्रिविमीय गोलीय निर्देशांकों हेतु)।

Explanation not matching this question? क्या व्याख्या इस प्रश्न से मेल नहीं खा रही? Report

130

ICAR PG Statistical Sciences 09 Jul 2023Shift 1

MCQ (Single Correct) +4 −1

The solution of the differential equation $\cos(x+y)dx + [3y^2 + 2y + \cos(x+y)]dy = 0$ is given by :

अवकल समीकरण $\cos(x+y)dx + [3y^2 + 2y + \cos(x+y)]dy = 0$ का हल होगा :

$\sin(x+y) + y^3 + y^2 = C$ $\sin(x+y) + y^3 + y^2 = C$ $\sin(x-y) + y^3 + y^2 = C$ $\sin(x-y) + y^3 + y^2 = C$ $\sin(x+y) + x^3 + x^2 = C$ $\sin(x+y) + x^3 + x^2 = C$ $\sin(x-y) + x^3 + x^2 = C$ $\sin(x-y) + x^3 + x^2 = C$

The given differential equation is of the form $M\,dx + N\,dy = 0$ with:
$M = \cos(x+y)$,
$N = 3y^2 + 2y + \cos(x+y)$.
Check for exactness:
$\frac{\partial M}{\partial y} = -\sin(x+y)$,
$\frac{\partial N}{\partial x} = -\sin(x+y)$.
Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.
The solution is obtained by:
$\int M\,dx$ (treating $y$ as constant) $+ \int (\text{terms in } N \text{ free from } x)\,dy = C$.
$\int \cos(x+y)\,dx = \sin(x+y)$.
Terms of $N$ free from $x$: $3y^2 + 2y$, so $\int (3y^2 + 2y)\,dy = y^3 + y^2$.
Combining gives:
$\sin(x+y) + y^3 + y^2 = C$.

Detailed analysis of options:

$\sin(x+y) + y^3 + y^2 = C$ is correct.
Has $x-y$ instead of $x+y$.
Replaces $y^3 + y^2$ with $x^3 + x^2$.
Has $x-y$ and $x$ terms.

दिया गया अवकल समीकरण $M\,dx + N\,dy = 0$ के रूप में है, जहाँ:
$M = \cos(x+y)$,
$N = 3y^2 + 2y + \cos(x+y)$.
यथार्थता (exactness) की जाँच:
$\frac{\partial M}{\partial y} = -\sin(x+y)$,
$\frac{\partial N}{\partial x} = -\sin(x+y)$.
चूँकि $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ है, अतः यह एक यथार्थ अवकल समीकरण (exact differential equation) है।
व्यापक हल:
$\int M\,dx$ ($y$ को अचर मानकर) $+ \int (N \text{ के वे पद जिनमें } x \text{ न हो})\,dy = C$.
$\int \cos(x+y)\,dx = \sin(x+y)$.
$N$ के $x$-रहित पद $3y^2 + 2y$ हैं, जिनका समाकल $y^3 + y^2$ है।

अतः अभीष्ट हल: $\sin(x+y) + y^3 + y^2 = C$. विकल्पों का विस्तृत विश्लेषण: