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प्रश्न
Find the general solution of the following differential equation :
`(1+y^2)+(x-e^(tan^(-1)y))dy/dx= 0`
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उत्तर
Given:
`(1+y^2)+(x-e^(tan^(-1)y))dy/dx= 0`
Let tan−1y=t
⇒y=tant
`=>dy/dx=sec^2tdt/dx`
Therefore, the equation becomes
(1+tan2t)+(x−et)sec2t `dt/dx=0`
`=>sec^2t+(x-e^t)(sec^2t)dt/dx=0`
`=>1+(x-e^t)dt/dx=0`
`=>(x-e^t)dt/dx=-1`
`=>x-e^t=dx/dt`
`=>dx/dt+1.x=e^t`
If =e∫1.dt
= et
`:. e^t.(dx/dt+1.x)=e^t.e^t`
`=>d/dt(xe^t)=e^(2t)`
Integrating both the sides, we get
`xe^t=inte^(2t)dt`
`=>xe^t=1/2e^(2t)+C " ....(1)"`
Substituting the value of t in (1), we get
`xe^(tan^(1))y=1/2e^(2tan^(-1)y)+C_1`
`=>e^2tan^(-1y)=2xe^(tan^1y)+C`
It is the required general solution.
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