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Question

The solution of log $$\left [ \dfrac{dy}{dx} \right ]=3x+4y$$ given that $$y(0)=0$$


A
4e3x3e4y=1
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B
3e3x+4e4y=7
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C
4e3x+3e4y=7
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D
3e3x+4ey=1
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Solution

The correct option is D $$4e^{3x}+3e^{-4y}=7$$
$$logy^{1}=3x+4y$$
$$\Rightarrow \dfrac{dy}{dx}=e^{3x+4y}$$
$$\Rightarrow \int {e^{-4y}dy=\int {e^{3x}dx+e}}$$
$$\dfrac{e^{-4y}}{-4}=\dfrac{e^{3x}}{3}+c$$
$$3e^{-4y}+4e^{3x}=c$$

Mathematics

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