The solution of the differential equation (d³ y/dx³) + (d² y/dx²) + 4(dy/dx) + 4y = 0 is given by : / अवकल समीकरण (d³ y/dx³) + (d² y/dx²) + 4(dy/dx) + 4y = 0 का हल है:

y = C₁ e^{-x} + C₂ 2x + C₃ 2x

y = C₁ e^x + C₂ 2x + C₃ 2x

y = C₁ e^{-x} + C₂ 2x + C₃ 2x

y = C₁ e^{2x} + C₂ 2x + C₃ 2x

y = C₁ e^{-2x} + C₂ 2x + C₃ 2x

Mathematics & Numerical Analysis Practice Questions (Q131-135) - Statistical Sciences

131

ICAR PG Statistical Sciences 09 Jul 2023Shift 1

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 :

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

132

ICAR PG Statistical Sciences 09 Jul 2023Shift 1

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 :

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

133

ICAR PG Statistical Sciences 14 Jul 2023Shift 1

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।

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

134

ICAR PG Statistical Sciences 14 Jul 2023Shift 1

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:

$x^3 y^5 (2.5 + c x^2) = 1$ सही हल है।
$x$ और $y$ की घातें बदली हुई हैं।
कोष्ठक में $x^2$ के स्थान पर $y^2$ है।
पूर्णतः त्रुटिपूर्ण है।

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

135

ICAR PG Statistical Sciences 14 Jul 2023Shift 1

MCQ (Single Correct) +4 −1

The solution of the differential equation $\frac{d^3 y}{dx^3} + \frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 0$ is given by :

अवकल समीकरण $\frac{d^3 y}{dx^3} + \frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 0$ का हल है:

$y = C_1 e^x + C_2 \cos 2x + C_3 \sin 2x$ $y = C_1 e^x + C_2 \cos 2x + C_3 \sin 2x$ $y = C_1 e^{-x} + C_2 \cos 2x + C_3 \sin 2x$ $y = C_1 e^{-x} + C_2 \cos 2x + C_3 \sin 2x$ $y = C_1 e^{2x} + C_2 \cos 2x + C_3 \sin 2x$ $y = C_1 e^{2x} + C_2 \cos 2x + C_3 \sin 2x$ $y = C_1 e^{-2x} + C_2 \cos 2x + C_3 \sin 2x$ $y = C_1 e^{-2x} + C_2 \cos 2x + C_3 \sin 2x$