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The solution of the equation $x \frac{d y}{d x}+3 y=x$ is
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Verified Answer
The correct answer is:
$x^3 y=\frac{x^4}{4}+c$
$x \frac{d y}{d x}+3 y=x \quad \Rightarrow \quad \frac{d y}{d x}+\frac{3 y}{x}=1$
It is in the form of $\frac{d y}{d x}+P y=Q$
So, I.F. $=e^{\int P d x}=e^{3 \int \frac{1}{x} d x}=e^{3 \log x}=x^3$
Hence required solution is
$y x^3=\int x^3 1 d x=\frac{x^4}{4}+c \Rightarrow y x^3-\frac{x^4}{4}+c$
It is in the form of $\frac{d y}{d x}+P y=Q$
So, I.F. $=e^{\int P d x}=e^{3 \int \frac{1}{x} d x}=e^{3 \log x}=x^3$
Hence required solution is
$y x^3=\int x^3 1 d x=\frac{x^4}{4}+c \Rightarrow y x^3-\frac{x^4}{4}+c$
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