Physicsss, where you at? Try and follow this one and then I have one for you to try if you want. The technique Qbert suggested is new to me so this is my first attempt at using such and applying it to an IVP:
For the equation:
Py^{''}+Qy^{'}+Ry=0
If it's exact we can immediately write:
(Py^{'})^{'}+(y\int R)^{'}=0
Example:
x^2y^{''}+(2x+\frac{1}{4}x^4)y^{'}+x^3y=0;\quad y(0.1)=0,\quad y^{'}(0.1)=1
Noting that the middle function is the sum of the derivative of the first function and the antiderivative of the last function, we can then say:
(x^2y^{'})^{'}+(\frac{1}{4}x^4y)^{'}=0
Taking antiderivatives:
\int (x^2y^{'})^{'}+\int(\frac{1}{4}x^4y)^{'}=\int 0
Yielding:
x^2y^{'}+1/4x^4y=c_1
Setting up this equation in the standard form to determine the integrating factor, we divide by x^2
y^{'}+1/4x^2y=c_1x^{-2}
The integrating factor is:
\sigma=Exp\left(\int 1/4 x^2dx\right)=e^{x^3/12}
Multiplying the ODE by \sigma converts the LHS to an exact differential:
d(e^{x^3/12}y)=c_1x^{-2}e^{x^3/12}
Integrating:
\int_{\xi=0.1,\zeta=0}^{\xi=x,\zeta=y} d(e^{\xi^3/12}\zeta)=c_1\int_{0.1}^{x}t^{-2}e^{t^3/12}dt
This yields:
e^{x^3/12}y(x)=c_1\int_{0.1}^x t^{-2}e^{t^3/12}dt
or:
y(x)=c_1e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt
So how do we find the constant? What about differentiating y(x) using Leibnitz's rule for the integral:
y^{'}(x)=c_1\left[e^{-x^3/12}\frac{d}{dx}\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)+\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)\frac{d}{dx}e^{-x^3/12}\right]
I know it's messy but try to follow it alright?
Now, we know from the IVP that the derivative at 0.1 is 1. Now look at the second sum. When we insert x=0.1 into the integral, that will be zero right? We're left with then:
y^{'}(x)=c_1e^{-x^3/12}\left(x^{-2}e^{x^3/12}\right)=\frac{c_1}{x^2}
Solving:
y^{'}(0.1)=\frac{c_1}{(0.1)^2}=1
We have c_1=0.01
Thus the solution is:
y(x)=0.01e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt
Now, there's nothing wrong with leaving it that way. Perfectly fine. You're comfortable with functions like Sin(x), Tan(x) right? What about a Sal(x)?
Let:
Sal(x)=\int_{0.1}^x t^{-2}e^{t^3/12}dt
Then the solution is simply:
y(x)=0.01e^{-x^3/12}Sal(x)
No different than using Sin(x) ok?
A plot is attached.
Now try this one:
6y^{''}+(\cos x) y^{'}-(\sin x) y=0; \quad y(0)=0,\quad y^{'}(0)=1
I have no idea how it will turn out.
