Solve : $\log\left(\frac{dy}{dx}\right)=ax+by$

Solve : log((dy)/(dx))=ax+by

Question 29 of 43

\log\left(\frac{dy}{dx}\right)=ax+by

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2024Subjective· 2mImportance★★★★★

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$\frac{dy}{dx}=e^{ax}e^{by}$ ; separating and integrating gives $\frac{e^{ax}}{a}+\frac{e^{-by}}{b}=c$ .

From $\log\left(\dfrac{dy}{dx}\right)=ax+by$ ,

$\frac{dy}{dx}=e^{ax+by}=e^{ax}\,e^{by}.$

Separate the variables:

$e^{-by}\,dy=e^{ax}\,dx.$

Integrate both sides:

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set MARCH1 markMCQ
$\frac{dy}{dx} = \cos x$
(a) $y = \cos x + c$ , c is an arbitrary constant
(b) $y = \sin x + 1$
(c) $y = \sin x + c$ , c is an arbitrary constant
(d) $y = \sin x - 2$
›Reveal solutionSolution

$\int dy = \int \cos x\,dx$ gives the general solution $y=\sin x + c$ .

The equation $\dfrac{dy}{dx}=\cos x$ is directly variable-separable:

$dy = \cos x\,dx.$

Integrating both sides:

$\int dy = \int \cos x\,dx \;\Rightarrow\; y = \sin x + c,$

…

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CBSE 2024Set MARCH1 markMCQ
$\frac{dy}{dx}=\cos x$
(a) $y=\sin x-2$
(b) $y=\cos x+c$ , $c$ is an arbitrary constant
(c) $y=\sin x+1$
(d) $y=\sin x+c$ , $c$ is an arbitrary constant
›Reveal solutionSolution

$y=\int\cos x\,dx=\sin x+c$ .

Separating and integrating $\dfrac{dy}{dx}=\cos x$ :

$y=\int\cos x\,dx=\sin x+c,$

where $c$ is an arbitrary constant (a first-order equation's general solution must contain one arbitrary constant). Options with fixed …

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CBSE 2022Set MARCH1 markMCQ
$\frac{dx}{dy} = f\left(\frac{x}{y}\right)$
(a) $y = v$
(b) $x = vy$
(c) $x = v$
(d) $y = vx$
›Reveal solutionSolution

Use $x = vy$ .

When a differential equation has the homogeneous form $\dfrac{dx}{dy} = f\!\left(\dfrac{x}{y}\right)$ , the right side depends only on the ratio $\dfrac{x}{y}$ . The standard technique is to set that ratio equal to a new variable:

$x = vy \quad\Rightarrow\quad \frac{x}{y} = v, \qquad \frac{dx}{dy} = v + y\frac{dv}{dy}.$

…

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Q.Solve : $\log\left(\frac{dy}{dx}\right)=ax+by$

Concept understanding — Variable-Separable Method

Q.Solve : log⁡(dydx)=ax+by\log\left(\frac{dy}{dx}\right)=ax+bylog(dxdy​)=ax+by

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Q.Solve : $\log\left(\frac{dy}{dx}\right)=ax+by$

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