General solution of the differential equation | vedclass

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Question

(C) Given differential equation is $(y^3+y)(x^2+1) dy = x(y^4+2y^2) dx$. Separating the variables,we get: $\frac{y^3+y}{y^4+2y^2} dy = \frac{x}{x^2+1}

Vedclass · Accepted Answer

$y^2(y^2+2) = C(x^2+1)^2$

Vedclass · Answer

(C) Given differential equation is $(y^3+y)(x^2+1) dy = x(y^4+2y^2) dx$. Separating the variables,we get: $\frac{y^3+y}{y^4+2y^2} dy = \frac{x}{x^2+1} dx$. Multiply the numerator and denominator of the left side by $2$: $\frac{1}{2} \int \frac{2y^3+2y}{y^4+2y^2} dy = \int \frac{x}{x^2+1} dx$. Let $u = y^4+2y^2$,then $du = (4y^3+4y) dy = 2(2y^3+2y) dy$,so $(2y^3+2y) dy = \frac{1}{2} du$. Substituting this: $\frac{1}{2} \int \frac{1}{2u} du = \int \frac{x}{x^2+1} dx$. $\frac{1}{4} \ln|y^4+2y^2| = \frac{1}{2} \ln|x^2+1| + \ln|C_1|$. Multiply by $4$: $\ln|y^4+2y^2| = 2 \ln|x^2+1| + 4 \ln|C_1| = \ln|(x^2+1)^2| + \ln|C_1^4|$. Taking exponential on both sides: $y^4+2y^2 = C(x^2+1)^2$,where $C = C_1^4$. $y^2(y^2+2) = C(x^2+1)^2$.

General solution of the differential equation $(y^3+y)(x^2+1) dy = (xy^4+2y^2x) dx$ is (where $C$ is a constant of integration.)

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