Solution of the differential equation $l

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Solution of the differential equation $log(\frac{dy}{dx})=3x+4y $ given that $y=0$ when $x=0 $ is :

Options:

$3e^{3x}+4e^{-4t}+7=0$

$-4e^{3x}+3e^{-4y}+7=0$

$4e^{3x}+3e^{-4y}-7=0$

$-3e^{3x}-4e^{-4y}+7=0$

Correct Answer:

$4e^{3x}+3e^{-4y}-7=0$

Explanation:

$\log(\frac{dy}{dx})=3x+4y⇒\frac{dy}{dx}=e^{3x}e^{4y}$

so $\int e^{-4y}dy=\int e^{3x}dx$

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

so $4e^{3x}+3e^{-4y}+c'=0$