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The solution of the differential equation dydx+yxlogy=yx2(logy2) is
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[c] Divide the equation of by y(logy)2 1y(logy)2dydx+1logy.1x=1x2 Put 1logy=z⇒−1y(logy)2dydx=dzdx Thus, we get, −dzdx+1x.z=1x2, linear in z ⇒dzdx+(−1x)z=−1x2; I.F. =e−∫1xdx=e−logx=1x ∴ The solution is, z(1x)=∫−1x2(1x)dx+c ⇒1logy(1x)=−x−2−2+c⇒x=logy(cx2+12)
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Question Text | The solution of the differential equation dydx+yxlogy=yx2(logy2) is |
Answer Type | Text solution:1 |
Upvotes | 150 |