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  • Find the general solution of the differential equation (1 + y^{2}) + (x - etan-1y)frac{dy}{dx}=0

Find the general solution of the differential equation (1 + y^{2}) + (x - e^{\tan^{-1}y})\frac{dy}{dx}=0

Answers (1)

Given

(1+y^{2})+(x-e^{\arctan y})\frac{dy}{dx}=0

To find: Solution of the given differential equation

Rewrite the given equation as,

(1+y^{2})\frac{dx}{dy}+x-e^{\arctan y}=0\\ \frac{dx}{dy}+\frac{x}{(1+y^{2})}=\frac{e^{\arctan y}}{(1+y^{2})}

It is a first order differential equation

Comparing it with

\frac{dx}{dy}+p(y)x=q(y)\\\\ p(y)=\frac{1}{(1+y^{2})}\\\\ q(y)=\frac{e^{\arctan y}}{(1+y^{2})}

Calculating Integrating Factor

IF=e^{\int p(y)dy}\\ IF=e^{\int \frac{1}{1+y^{2}}dy}\\ Formula: \int \frac{1}{(1+y^{2})}dy=\arctan y\\IF=e^{\arctan y}

Hence the solution of the given differential equation is

x.(IF)=\int q(y).(IF)dy\\ x.(e^{\arctan y})=\int \frac{e^{\arctan y}}{(1+y^{2})}(e^{\arctan y })dy\\$ Assume $ (e^{\arctan y })=t

Differentiate on both the sides

\frac{e^{\arctan y }}{(1+y^{2})}dy=dt\\ x.t = \int tdt \\x.t = \frac{t^2}{2}+c \\$Substituting$ \; t\\ x.(e^{\tan^{-1}y})=\frac{e^{2\tan^{-1}y}}{2}+c

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