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`        solve the differential equation:x(1-x^2)dy/dx +(2x^2-1)y=ax^3 `
8 months ago

```							dy/dx+(2x^2−1)/(x(1−x^2))y=x^(2)/(1−x^2)P(x)=(2x^2−1)/(x(1−x^2)) and Q(x)=x^(2)/(1−x^2)I.F.=e^[ ∫{-(1-x^2-x^2)/(x(1-x^2))}dx]I.F.=e^[ ∫{-(1/x-x/(1-x^2))}dx]I.F.=e^[ {-ln(x)-1/2ln(1-x^2)}]I.F.=e^[ ln{(x)*(1-x^2)^(1/2)}^(-1)]I.F.=1/ {(x)*(1-x^2)^(1/2)}then solutiony(I.F.)=∫ Q(x)*(I.F.)dxy/ {(x)*(1-x^2)^(1/2)}=∫x/(1−x^2)^(3/2)dxy/ {(x)*(1-x^2)^(1/2)}=1/(1−x^2)^(1/2)+c
```
8 months ago
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