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Help with differentials?
How would you find the general solution of
(1-x^2) y' - xy = 1
I tried by using the reverse of differentiation by parts but I got nowhere.
1 Answer
- NiallLv 76 years agoFavorite Answer
This is a linear equation, put it in the form of y' + p(x)y = g(x):
y' - x / (1 - x^2) * y = 1 / (1 - x^2)
Now we find the integrating factor, which is I = e^( ∫ P(x))
I = e^[ ∫ -x / (1 - x^2) * dx]
I = e^[1/2 * ln(1 - x^2)]
I = (1 - x^2)^1/2
Multiply throguh by the integrating factor:
(1 - x^2)^1/2 * y' - x / (1 - x^2)^1/2 * y = 1 / (1 - x^2)^1/2
After we multiply through by the integrating factor, the left side becomes the derivative of the product of y and the integrating factor (which you can check with the product rule):
d/dx [(1 - x^2)^1/2 * y] = 1 / (1 - x^2)^1/2
Integrate both sides:
(1 - x^2)^1/2 * y = arcsin(x) + C
y = [arcsin(x) + C] / (1 - x^2)^1/2