Differential equations
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Solve
(1-x^2)^(1/2)y'+1+y^2=0
xy(1+x^2)y'-(1+y^2)=0
xyy'=1+x^2+y^2+x^2y^2
sinx(e^y + 1)dx=e^y(1+cosx)dy, Y(0)=0
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Solution Preview
1. (1-x^2)^(1/2)y'+1+y^2=0 implies that
y'/(1+y^2)=-1/((1-x^2)^(1/2) ) (1)
We have the formula
The definite integral of
1/((a^2+u^2) is 1/a* [ tan^(-1)(u/a)] +C
The definite integral of
1/((a^2-u^2)^(1/2) is sin^(-1)(u/a)+C.
Integrate both sides of (1) to get ( with a=1)
1/[ tan^(-1)(y)] = - sin^(-1)(x)+C.
Thus
tan(y)= 1/[- ...
Solution Summary
This showsh ow to solve several different differential equations.
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