The solution of the differential equation x(dy/dx) + y = x³ y⁶...

Q.6

ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper (Q.106)

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:

$\sin(x+y) + y^3 + y^2 = C$ सही उत्तर है।
इसमें $x-y$ है।
इसमें $y$ के स्थान पर $x$ लिखा है।
गलत है।

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

Q.7

ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper (Q.107)

MCQ (Single Correct) +4 −1

The solution of the differential equation $xy'' + y' = 4x$ is given by :

अवकल समीकरण $xy'' + y' = 4x$ का हल होगा :

$y = x^3 + C_1 \ln x + C_2$ $y = x^3 + C_1 \ln x + C_2$ $y = x^2 + C_1 \ln x + C_2$ $y = x^2 + C_1 \ln x + C_2$ $y = e^x + C_1 \ln x + C_2$ $y = e^x + C_1 \ln x + C_2$ $y = e^{-x} + C_1 \ln x + C_2$ $y = e^{-x} + C_1 \ln x + C_2$

Q.8

ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper (Q.108)

MCQ (Single Correct) +4 −1

The solution of the differential equation $\frac{d^2x}{dt^2} + 6\frac{dx}{dt} + 9x = 0$ is given by :

अवकल समीकरण $\frac{d^2x}{dt^2} + 6\frac{dx}{dt} + 9x = 0$ का हल होगा :

$x = (C_1 + C_2 t)e^{3t}$ $x = (C_1 + C_2 t)e^{3t}$ $x = (C_1 + C_2 t)e^{-3t}$ $x = (C_1 + C_2 t)e^{-3t}$ $x = (C_1 + C_2 t)e^{5t}$ $x = (C_1 + C_2 t)e^{5t}$ $x = (C_1 + C_2 t)e^{-5t}$ $x = (C_1 + C_2 t)e^{-5t}$

Q.9

ICAR AIEEA PG Statistical Science 14 July 2023 Shift-1 Paper (Q.106)

MCQ (Single Correct) +4 −1

The solution of the differential equation $(1 + y e^{xy})dx + (2y + x e^{xy})dy = 0$ is given by :

अवकल समीकरण $(1 + y e^{xy})dx + (2y + x e^{xy})dy = 0$ का हल है:

$x + e^{xy} + y^2 = C$ $x + e^{xy} + y^2 = C$ $y + e^{xy} + x^2 = C$ $y + e^{xy} + x^2 = C$ $e^x + xy + e^y = C$ $e^x + xy + e^y = C$ $e^x - xy + e^y = C$ $e^x - xy + e^y = C$

Solution of exact first-order differential equations: Standard form: $M dx + N dy = 0$, where: $M = 1 + y e^{xy} \implies \frac{\partial M}{\partial y} = e^{xy} + xy e^{xy}$. $N = 2y + x e^{xy} \implies \frac{\partial N}{\partial x} = e^{xy} + xy e^{xy}$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. The general solution is $\int M dx + \int (\text{terms of } N \text{ not containing } x) dy = C$: $\int (1 + y e^{xy}) dx = x + e^{xy}$. $\int (2y) dy = y^2$. Combining gives $x + e^{xy} + y^2 = C$. Option breakdown:

$x + e^{xy} + y^2 = C$ is the exact potential function.
Swaps the roles of $x$ and $y$.
Incorrect algebraic terms.
Incorrect algebraic terms.

यथातथ (Exact) अवकल समीकरण का हल:
मानक रूप $M dx + N dy = 0$, जहाँ:
$M = 1 + y e^{xy} \implies \frac{\partial M}{\partial y} = e^{xy} + xy e^{xy}$.
$N = 2y + x e^{xy} \implies \frac{\partial N}{\partial x} = e^{xy} + xy e^{xy}$.
चूँकि $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ है, अतः यह एक यथातथ अवकल समीकरण है।
व्यापक हल:
$\int M dx$ ($y$ को अचर मानकर) $+ \int (N$ के वे पद जिनमें $x$ न हो) $dy = C$:
$\int (1 + y e^{xy}) dx = x + e^{xy}$.
$\int 2y \, dy = y^2$.

अतः अभीष्ट हल $x + e^{xy} + y^2 = C$ है। विकल्पों का विश्लेषण:

$x + e^{xy} + y^2 = C$ सही हल है।
$x$ और $y$ के पद बदले हुए हैं।
गलत पद हैं।
गलत पद हैं।

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Q.10

ICAR AIEEA PG Statistical Science 14 July 2023 Shift-1 Paper (Q.107)

MCQ (Single Correct) +4 −1

The solution of the differential equation $x\frac{dy}{dx} + y = x^3 y^6$ is given by :

अवकल समीकरण $x\frac{dy}{dx} + y = x^3 y^6$ का हल है:

$x^3 y^5 (2.5 + c x^2) = 1$ $x^3 y^5 (2.5 + c x^2) = 1$ $x^5 y^3 (2.5 + c x^2) = 1$ $x^5 y^3 (2.5 + c x^2) = 1$ $x^3 y^5 (2.5 + c y^2) = 1$ $x^3 y^5 (2.5 + c y^2) = 1$ $x^5 y^3 (2.5 + c y^2) = 1$ $x^5 y^3 (2.5 + c y^2) = 1$

Solution of Bernoulli differential equation: Divide both sides by $x y^6$: $y^{-6} \frac{dy}{dx} + \frac{1}{x} y^{-5} = x^2$. Let $v = y^{-5} \implies \frac{dv}{dx} = -5 y^{-6} \frac{dy}{dx} \implies y^{-6} \frac{dy}{dx} = -\frac{1}{5} \frac{dv}{dx}$. Substituting into the equation: $-\frac{1}{5} \frac{dv}{dx} + \frac{1}{x} v = x^2 \implies \frac{dv}{dx} - \frac{5}{x} v = -5 x^2$. Integrating Factor (I.F.): $\text{I.F.} = e^{\int -\frac{5}{x} dx} = e^{-5\ln x} = x^{-5}$. Solution for $v$: $v \cdot x^{-5} = \int (-5 x^2)(x^{-5}) dx = -5 \int x^{-3} dx = -5 \left( \frac{x^{-2}}{-2} \right) + c = 2.5 x^{-2} + c$. Multiplying by $x^5$: $v = 2.5 x^3 + c x^5 = x^3 (2.5 + c x^2)$. Since $v = y^{-5} = \frac{1}{y^5}$: $\frac{1}{y^5} = x^3 (2.5 + c x^2) \implies x^3 y^5 (2.5 + c x^2) = 1$. Option breakdown:

$x^3 y^5 (2.5 + c x^2) = 1$ is the exact solution.
Swaps exponents of $x$ and $y$.
Has $y^2$ inside the parentheses instead of $x^2$.
Swaps exponents and has $y^2$ inside parentheses.

बरनौली अवकल समीकरण का हल:
समीकरण के दोनों पक्षों को $x y^6$ से भाग देने पर:
$y^{-6} \frac{dy}{dx} + \frac{1}{x} y^{-5} = x^2$.
माना $v = y^{-5} \implies \frac{dv}{dx} = -5 y^{-6} \frac{dy}{dx}$.
समीकरण में रखने पर:
$\frac{dv}{dx} - \frac{5}{x} v = -5 x^2$.
समाकलन गुणक (I.F.):
$\text{I.F.} = e^{\int -\frac{5}{x} dx} = x^{-5}$.
$v$ का हल:
$v \cdot x^{-5} = \int -5 x^{-3} dx = 2.5 x^{-2} + c$.
$x^5$ से गुणा करने पर:
$v = x^3 (2.5 + c x^2)$.
$v = y^{-5}$ रखने पर:
$\frac{1}{y^5} = x^3 (2.5 + c x^2) \implies x^3 y^5 (2.5 + c x^2) = 1$.

विकल्पों का विश्लेषण: