General solution of the differential equation dy/dx = e^(ax+by) is: (where C is an arbitrary constant) Option 1: e^ax + e^(-by) = c Option 2: e^(ax+by) = c Option 3: be^ax + ae^(-by) = c Option 4: ae^ax - be^(-by) = c

Answer: be^ax + ae^(-by) = c. Solution: Write dy/dx = e^axe^by and separate: e^(-by)dy = e^axdx. Integrating gives -(e^(-by))/b = e^ax/a + k. Multiplying by -ab gives be^ax + ae^(-by) = c.

Calculus

Differential Equations

Medium

common

General solution of the differential equation $\dfrac{dy}{dx} = e^{ax+by}$ is: (where C is an arbitrary constant)

$e^{ax} + e^{-by} = c$

$e^{ax+by} = c$

$be^{ax} + ae^{-by} = c$

$ae^{ax} - be^{-by} = c$

✅ Correct Option: 3

Write $\frac{dy}{dx} = e^{ax}e^{by}$ and separate: $e^{-by}dy = e^{ax}dx$ . Integrating gives $-\frac{e^{-by}}{b} = \frac{e^{ax}}{a} + k$ . Multiplying by $-ab$ gives $be^{ax} + ae^{-by} = c$ .

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The equation of the curve that passes through $(-1, 1)$ and the slope of line touching to the curve at any point $(x, y)$ is $\frac{3x^2}{y}$ , is:

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The degree of the differential equation $\left[1 + \left(\frac{dy}{dx}\right)\right]^3 = \left(\frac{d^2 y}{dx^2}\right)^2$ is :

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If the order and degree of the differential equation $y + \dfrac{dy}{dx} = \int y\,dx$ are m and n respectively, then (m - n) is equal to

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The general solution of the differential equation $x\,dx = (x^2+1)\,dy$ is (where C is an arbitrary constant)

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The differential equation representing the curve $y = e^{2x}(a + bx)$ , where a, b are arbitrary constants is

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Match the LIST-I with LIST-II

	LIST-I Differential equation		LIST-II Sum of the order and the degree of the differential equation
A.	$\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = 1$	I.	$2$
B.	$4 + \left(\dfrac{dy}{dx}\right)^4 = 7\left(\dfrac{d^2y}{dx^2}\right)^3$	II.	$3$
C.	$x^3\left(\dfrac{d^2y}{dx^2}\right)^2 + x\left(\dfrac{dy}{dx}\right)^4 + 1 = 0$	III.	$5$
D.	$\dfrac{dy}{dx} = x^4 e^{-3y}$	IV.	$4$

Choose the correct answer from the options given below:

General solution of the differential equation dydx=eax+by\dfrac{dy}{dx} = e^{ax+by}dxdy​=eax+by is: (where C is an arbitrary constant)

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General solution of the differential equation $\dfrac{dy}{dx} = e^{ax+by}$ is: (where C is an arbitrary constant)