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Find:(dy)/(dx) - 1/x y = xy2\(\frac{dy}{dx}-\frac1{x}y = xy^2\) |
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Answer» \(\frac{dy}{dx}-\frac1{x}y = xy^2\) ⇒ \(\frac1{y^2}\frac{dy}{dx}-\frac1{x}.\frac1{y}=x\)----(1) Let \(\frac1y=z\) Then \(-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}\) Then from (1), we obtain \(-\frac{dz}{dx}-\frac1{x}z=x\) ⇒ \(\frac{dz}{dx}+\frac1{x}z=-x\) \(\therefore\) I.F. = e\(\int\)pdx = e\(\int\) (1/x)dx = elog x = x \(\therefore\) Complete solution is y x I.F. = \(\int\)(I.F.). Q dx ⇒ y \(\times\) x = \(\int\)x \(\times\) -x dx ⇒ yx = -\(\int\)x2dx ⇒ yx = \(-\frac{x^3}3+C\) ⇒ y = \(-\frac{x^2}3+\frac{C}x\) Hence, solution of given differential equation is y = \(-\frac{x^2}3+\frac{C}x\) |
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