An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows .
In the given problem: `(du)/(dv)=uvsin(v^2)` , we may apply variable separable differential equation in a form of .
Divide both sides by "u" and cross-multiply dv to set it up as:
Apply direct integration: `int(du)/u=int vsin(v^2) dv.`
For the left sign, we follow the basic integration formula for logarithm:
`int (du)/u = ln|u|`
For the right side, we follow the basic integration formula for sine function:
Let: `w=v^2` then `dw = 2v*dv` or `(dw)/2 =v dv` .
The integral becomes:
`intvsin(v^2) dv= intsin(v^2) * vdv`
`= (1/2) int sin(w) dw`
Plug-in `w=v^2` on `-cos(w)/2+C` , we get:
Combing the results, we get the general solution of differential equation as:
`ln|u| = -cos(v^2)/2+C`
To solve for the arbitrary constant `(C)` , apply the initial condition `u(0)=1` on`ln|u| = -cos(v^2)/2+C` :
`ln|1| = -cos(0^2)/2+C`
`0 = -1/2+C`
`C = 0+1/2`
Plug-in ` C= 1/2` in `ln|u| = -cos(v^2)/2+C` , we get
`ln|u| = -cos(v^2)/2+1/2`
`u = e^(-cos(v^2)/2+1/2)`