A homegeneous first-order ode has the form:
where g(s) is a given function.
Solution Procedure
To solve this problem, we introduce the variable v(t)=y(t)/t and derive a new differential equation for v(t). We have y(t)=v(t)t. Hence by the product rule, y'(t)=tv'(t)+v(t). Substituting this into the original ode we have
Cleaning this up, we have
What have we accomplished? (*) is a separable ode! We now use the techniques for separable ode to solve this problem. After solving for v(t) we get y(t) using the formula y(t)=tv(t).
Example
ODE:
Note that y'=g(y/t) where
We make the substitution v(t)=y(t)/t or y(t)=v(t)t and have y'=v+tv'. Substituting this into the differential equation and using the fact that v=y/t, we have
Cleaning this up and solving for v' we have
This is a separable ode. Applying the method for separable ode we have:
where C is a constant. The v integral is arctan(v) and the t integral is log|t|. Hence we have
Taking the tangent of both sides, we have
Recall that v(t)=y(t)/t or y(t)=tv(t). Hence
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