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NCERT Exemplar · Q10

Q.Find the general solution of (x+2y3)dydx=y(x+2y^3)\frac{dy}{dx}=y.

Himachal HpboseShort· 3mImportance★★★★★
Appeared in past exams:AP EAPCET 2025· Set eng-2025-05-21-AN· 1mrewordedAP EAPCET 2024· Set eng-2024-05-21-FN· 1mexact
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The equation is linear in xx (not in yy). Treating xx as a function of yy gives dxdy−xy=2y2\dfrac{dx}{dy}-\dfrac{x}{y}=2y^2, whose general solution is x=y3+Cyx = y^3 + Cy.

Swap the roles of xx and yy. From (x+2y3)dydx=y(x+2y^3)\dfrac{dy}{dx}=y, use dydx=1dx/dy\dfrac{dy}{dx}=\dfrac{1}{dx/dy}:

dxdy=x+2y3y=xy+2y2.\frac{dx}{dy} = \frac{x+2y^3}{y} = \frac{x}{y} + 2y^2.

Standard linear form:

dxdy−1y x=2y2,P(y)=−1y,Q(y)=2y2.\frac{dx}{dy} - \frac{1}{y}\,x = 2y^2,\qquad P(y)=-\frac1y,\quad Q(y)=2y^2.

Integrating factor:

μ(y)=e∫−1y dy=e−log⁡∣y∣=1y.\mu(y) = e^{\int -\frac1y\,dy} = e^{-\log|y|} = \frac{1}{y}.

Multiply through and integrate. The left side becomes an exact derivative: …

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