Given Differential equation is,
dxdy+(1−x2x)y=x×y Divide by y on both sides of the above equation, then we get
y1dxdy+(1−x2x)y=x.............(A) Let y=t
Differentiate with respect to x
Then,
2y1dxdy=dxdt Then, the equation (A) becomes
2dxdt+(1−x2x)t=x divide by 2
dxdt+21(1−x2x)t=2x
This is the Linear Differential equation in the form dxdt+P(x)t=Q(x)
Integrating factor = I.F = e∫p(x)dx=e∫21(1−x2x)dx
=e41∫1−x22xdx=e−41ln(1−x2)=41−x21 Now the solution of the linear differential equation is
t×(I.F)=∫(I.F)×Q×dx+c
t×41−x21=∫41−x21×2x dx+c
41−x2t=−41∫41−x2−2xdx+c
use substitution method on right side Integration,
Let 1−x2=m then −2xdx=dm
41−x2t=−41∫4m1dm+c
41−x2t=−41×41+1m41+1+c
41−x2t=−44×5m45+c
41−x2t=−5(1−x2)45+cy(0)=1
41−01=−5(1−0)45+c
1=5−1+cc=1+51=56
So the solution of the equation is,
41−x2y=−5(1−x2)45+56
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Thanks a lot for help.