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Solve the differential equation : (tan^−1 y−x)dy=(1+y^2)dx. - CBSE (Science) Class 12 - Mathematics

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Question

Solve the differential equation : (tan1yx)dy=(1+y2)dx.

Solution

(tan1yx)dy=(1+y2)dx.

Dividing both sides by (1 + y2), we get

`((tan^(-1)y−x)dy)/(1+y^2)=dx.........(1)`

Let t=tan-1 Y

Differentiating both sides with respect to y, we have

`(dt)/dy=1/(1+y^2)`

`=>dt=1/(1+y^2)dy`

From (1), we have

`tdt−xdt=dx`

`⇒(t−x)dt=dx`

`=>dx/dt=t-x`

`therefore dx/dt+x=t......(2)`

Here`I.F.=e^(int(1)dt)=e^t`

Hence, the solution of the differential equation (2) is given by

x(I.F.)=( I.F.)tdt

`⇒xe^t=∫ e^t.tdt+C, `

where C is an arbitrary constant

`⇒xe^t=te^t−∫e^tdt+C`

`⇒xe^t=te^t−e^t+C`

`⇒x=t−1+Ce^(−t)`

Substituting t=tan1y, we have

`x=tan^(−1)y−1+Ce^(−tan^(−1))y`

This is the general solution of the given differential equation.

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Solution Solve the differential equation : (tan^−1 y−x)dy=(1+y^2)dx. Concept: General and Particular Solutions of a Differential Equation.
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