Example 1

Today, we will solve the differential equation $y'' - y = e^x$. The first step is to find the general solution of the associated homogeneous equation, which is $y'' - y = 0$. Can anyone tell me how we can derive the characteristic equation?

Isn't it $r^2 - 1 = 0$?

Correct! We factor that gives us roots $r = 1$ and $r = -1$. So the general solution to the homogeneous equation is...

Is it $y_h(x) = C_1 e^x + C_2 e^{-x}$?

Exactly! Great job. Now we have our first solution. Remember, we denote these solutions as $y_1$ and $y_2$. Now let’s move to the next step!

Next, we need to find the Wronskian, which will help us evaluate the coefficients for the particular solution. Does anyone remember the formula for the Wronskian?

Yes, it’s $W = y_1 y_2' - y_2 y_1'$.

That's right! Now can we define $y_1$ and $y_2$ to compute the Wronskian?

Sure, $W = e^x(-e^{-x}) - e^{-x}(e^x) = -2$.

Perfect! Now we have $W = -2$. It's essential for our next steps.

Now, onto calculating the derivatives for our particular solution! Who can recall the formulas for $u_1'$ and $u_2'$?

I remember! It’s $u_1' = -\frac{y_2 g(x)}{W}$ and $u_2' = \frac{y_1 g(x)}{W}$.

Exactly! Great memory! Now substituting in our functions, what do we get for $u_1'$?

So we have $u_1' = -\frac{e^{-x} imes e^x}{-2} = \frac{1}{2}$.

Excellent! Now let's find $u_2'$ using the same steps.

Now, to find $u_1$ and $u_2$, we need to integrate the derivatives we calculated.

So, that means $u_1 = \int \frac{1}{2} \, dx = \frac{x}{2}$ and $u_2 = -\int \frac{e^{2x}}{2} \, dx = -\frac{e^{2x}}{4}$?

Exactly, nice work! Now, how do we construct the particular solution?

We combine them: $y_p = u_1 y_1 + u_2 y_2 = \frac{x}{2} e^x - \frac{e^{2x}}{4} e^{-x}$.

That's correct! Now we can finalize it and write down the general solution.

Finally, let's write the general solution for our differential equation. Does anyone want to summarize what we have?

We have $y(x) = C_1 e^x + C_2 e^{-x} + \frac{x}{2} e^x - \frac{e^{x}}{4}$.

Excellent summary! This solution combines the homogeneous and particular solutions successfully.

I find it easier to follow the steps when we systematically break it down like this!

I'm glad to hear that! Remember, each step is important in the method of variation of parameters!