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Dy/dx+(2xtan-¹y-x³)(1+y²)=0 semister 2nd bsc

Answer»

Given,

\frac{dy}{dx}+(2x\tan^{-<klux>1</klux>}y-x^3)(1+y^2)

Dividing by (1 + y²)

\frac{1}{1+y^2}\frac{dy}{dx}+(2x\tan^{-1}y-x^3)=0\\\;\\\text{Putting}\;\tan^{ - 1}y=t\\\;\\\implies\frac{1}{1+y^2}\frac{dy}{dx}=\frac{dt}{dx}\\\;\\\textbf{On doing above substitutions,}\\\;\\\frac{dt}{dx}+2xt-x^3 = 0\\\;\\\frac{dt}{dx}+2xt=x^3

This is the Linear Diffrentiatial EQUATION in t.

\text{I.F.}=e^{\int{2x\,dx}}\\\;\\\text{I.F.}=e^{x^2}

Thus SOLUTION of this diffrentiatial equation will be,

t\times\text{I.F.}=\int{x^2\times \text{I.F.}\,dx}+c\\\;\\t\times e^{x^2}=\int{x^3e{x^2}\,dx}+c\\\;\\te^{x^2}=e^{x^2}(x^2-1)+c\\\;\\\textbf{Putting the value of t,}\\\;\\e^{x^2}\tan^{-1}y=e^{x^2}(x^2-1)+c

This is the REQUIRED General Solution of given Diffrentiatial Equation.


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