Linear homogeneous equations
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2nd order linear homogeneous differential equations 3
In the last video we had this second order linear homogeneous differential equation and we just tried it out the solution y is equal to e to the rx. And we figured out that if you try that out, that it works for particular r's. And those r's, we figured out in the last one, were minus 2 and minus 3. But it came out of factoring this characteristic equation. And watch the last video if you forgot how we got that characteristic equation. And we ended up with this general solution for this differential equation. And you could try it out if you don't believe me that it works. But what if we don't want the general solution, we want to find the particular solution? Well then we need initial conditions. So let's do this differential equation with some initial conditions. So let's say the initial conditions are-- we have the solution that we figured out in the last video. Let me rewrite the differential equation. So it was the second derivative plus 5 times the first derivative plus 6 times the function, is equal to 0. And the initial conditions we're given is that y of 0 is equal to 2. And the first derivative at 0, or y prime at 0, is equal to 3. What does y equal at the point 0, and what is the slope at 0-- at x is equal to 0-- and the slope is 3. So how do we use these to solve for c1 and c2? Well, let's just use the first initial condition. y of 0 is equal to 2, which is equal to-- essentially just substitute 0 in into this equation. So it's c1 times e to the minus 2 times 0, that's essentially e to the 0, so that's just 1. So it's c1 times 1, which is c1, plus c2 times e to the minus 3 times 0. This is e to the 0, so it's just 1. So plus c2. So the first equation we get when we substitute our first initial condition is essentially c1 plus c2 is equal to 2. Now let's apply our second initial condition that tells us the slope at x is equal to 0. So y prime is 0. So this is our general solution, let's take its derivative, and then we can use this. So y prime of x is equal to what? The derivative of this is equal to minus 2 c1 times e to the minus 2x. And what's the derivative of this? It's minus 3 c2 times e to the minus 3x. And now we can use our initial condition, y prime at 0. So when x is equal to 0, what's the right-hand side equal? It's minus 2 times c1 and then e to the minus 0, e to the 0, that's just 1. Minus 3 c2, and then once again x is 0, so e to the minus 3 times 0, that's just 1. So it's just 1 times minus 3 c2. And it tells us that when x is equal to 0, what does this whole derivative equal? Well, it equals 3, right? Y prime of 0 is equal to 3. So now we go back into your first year of algebra. We have two equations-- two linear equations with two unknowns-- and we could solve. Let me write them in a form that you're probably more used to. So the first one is c1 plus c2 is equal to 2. And the second one is minus 2 c1 minus 3 c2 is equal to 3. So what can we do? Let's multiply this top equation by 2. There's a ton of ways to solve this, but if you multiply the top equation times 2, you'll get-- and I'll do this is a different color, just so that it's changed-- I'm just multiplying the top one by 2, you get 2 c1 plus 2 c2 is equal to 4. And now we can add these two equations. Minus 2 c1 plus 2, those cancel out. So minus 3 plus 2, you get minus c2 is equal to 7. Or we could say that c2 is equal to minus 7. And now we can substitute back in here. We have c1 plus c2-- c2 is minus 7-- so minus 7, is equal to 9, or we know that c-- oh sorry, no I'm already confusing myself. My brain was getting ahead of myself. c1 plus c2, that's minus 7, is equal to 2, right? I'm just substituting back into this differential equation-- sorry, to this equation, not a differential it's just a simple linear equation-- and then we get c1 is equal to 9. And now we have our particular solution to the differential equation. So this was our general solution. We can just substitute our c1's and our c2's back in. We have our particular solution for those initial conditions. And I think that warrants a different color. So our particular solution is y of x is equal to c1, which we figured out is 9e to the minus 2x, plus c2-- well, c2 is minus 7-- minus 7e to the minus 3x. That is the particular solution to our original differential equation. And it might be a good exercise for you to actually test it out. This particular solution to this differential equation. I'll do another example in the next video. I'll see you soon.