Question
समीकरण $x\frac{{dy}}{{dx}} + 3y = x$ का हल है
यह $\frac{{dy}}{{dx}} + Py = Q$ रूप का है।
अत: $I.F.$ $ = {e^{\int_{}^{} {Pdx} }} = {e^3}^{\int_{}^{} {\frac{1}{x}dx} } = {e^{3\log x}} = {x^3}$
अत: अभीष्ट हल
$yx^3 = \int {x^3}\ 1\ dx = \frac{{x^4}}{{4}} + c$ ==> $y{x^3} = \frac{{{x^4}}}{4} + c$
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