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If inta^x ty(t)dt=x^2+y(x), then find...

If `int_a^x ty(t)dt=x^2+y(x),` then find `y(x)`

Text Solution

Verified by Experts

`int_(a)^(x) ty(t)dt=x^(2)+y(x)`
Differentiating both sides, w.r.t x, we get
`xy(x)=2x+y^(')(x)`
Hence, `(dy)/(dx)-xy=-2x`(linear)
`I.F. E^(int-xdx)=e^(-x^(2)//2)`
Thus, solution is `ye^(-x^(2)//2)=int-2xe^(-x^(2)//2) dx`
or `y=2+ce^(x^(2)//2)`
If x=a, then `a^(2)+y=0` or `y=-a^(2)`
Hence, `-a^(2)=2+ce^(-a^(2)//2)`
or `ce^(-a^(2)//2) =-(2+a^(2))`
or `c=-(2+a^(2))e^(a^(2)//2)`
or `y=2-(2+a^(2))e^(a^(2)-x^(2))//2`
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