Or an ext specifically, a second-order *linear* homogeneous differential equation with facility roots. Yeesh, its always a mouthful through diff eq. Five and, we"ll litter in one initial problem just for sharks and also goggles. The trouble goes choose this:

Find a real-valued equipment to the initial value problem (y""+4y=0), through (y(0)=0) and also (y"(0)=1). Your solution have to be real-valued or you will certainly not receive complete credit!

If you"ll recall, the actions for solving a second-order homogeneous diff. Eq. Room as follows:

Write under the characteristics equation.Find the roots of the properties equation.Use the root to create down the 2 exponential communication solutions.Create a general solution utilizing a linear mix of the 2 basis solutions.You are watching: Find the real-valued solution to the initial value problem

For step 1, we just take ours differential equation and replace (y"") with (r^2), (y") with (r), and (y) with 1. Simple enough:

For action 2, we resolve this quadratic equation to gain two roots. The roots are going come be complicated numbers, but that"s ok:

Step 3 tells us to create down the resulting basis remedies using our 2 roots:

When we get to action 4 though, we have actually a problem. We"re supposed to find a *real-valued* solution. Clearly, noþeles that has (i) in it is no real-valued. So, room we SOL on this one? possibly not. It transforms out that we have the right to make a real-valued systems out of 2 non-real basis remedies using 2 tricks (and isn"t diff. Eq. All around tricks)?

Our first trick comes to us every the method from Switzerland via one Leonhard Euler (pronounced "Oiler"). The called, naturally enough, Euler"s Formula, and also tells us just how we deserve to turn a complicated exponential duty into a facility trigonometric function. Don"t worry about why it works, just know the the formula is:

Pretty cool, huh? The various other trick we"re walk to usage is the truth that any linear combination of services to a provided linear differential equation is **also** a solution. In various other words, we have the right to slice and dice our services (by multiplying them by constants and including them together) come get more solutions.

Obviously, we want to carry out our slicing and also dicing in a means that allows us end up with a real-valued solution. But, exactly which means is not so obvious. Therefore I"ll show you, and you can more or less rely on this technique for similar types that problems. We"re going to define two new solutions, one by including our basis solutions, and also one by subtracting them:

Using Euler"s Formula, we can rewrite those exponential basis remedies in regards to trig functions. Recall that cosine is one even duty and sine is an odd function, therefore we can do a bit of simplification with the an unfavorable root:

Plugging that into the equations for (y_3) and (y_4), we deserve to simplify them and also get two nice, clean, pretty new solutions:

Now, we can make a linear combination out of *those* services to get our basic solution:

Ah, you can say, however what about that (i) in the 2nd term? Well notice that now that it has been safely moved out to the former of the term, we deserve to simply define new consistent coefficients because that the terms. After ~ all, (i) may be non-real, yet its tho a constant. So, us define:

and get:

I know, cheap trick. But it works! now that we have our real-valued, general solution, we can use the initial conditions to discover (C_1) and (C_2) and get the equipment to our initial value problem:

There you have it.

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As you might have meant from other experiences you"ve had actually with differential equations, there space a collection of not-so-obvious tricks that you have to use to solve the problem. Your ideal bet exam-wise is to simply commit this tricks to memory, and also practice applying them over and also over to similar problems from her textbook.