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In the last video we had this second order linear

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homogeneous differential equation and we just tried it

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out the solution y is equal to e to the rx.

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And we figured out that if you try that out, that it works

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for particular r's.

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And those r's, we figured out in the last one, were minus 2

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and minus 3.

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But it came out of factoring this characteristic equation.

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And watch the last video if you forgot how we got that

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characteristic equation.

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And we ended up with this general solution for this

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differential equation.

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And you could try it out if you don't

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believe me that it works.

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But what if we don't want the general solution, we want to

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find the particular solution?

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Well then we need initial conditions.

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So let's do this differential equation with some initial

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conditions.

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So let's say the initial conditions are-- we have the

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solution that we figured out in the last video.

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Let me rewrite the differential equation.

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So it was the second derivative plus 5 times the

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first derivative plus 6 times the function, is equal to 0.

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And the initial conditions we're given is that y of 0 is

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equal to 2.

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And the first derivative at 0, or y prime at

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0, is equal to 3.

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What does y equal at the point 0, and what is the slope at

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0-- at x is equal to 0-- and the slope is 3.

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So how do we use these to solve for c1 and c2?

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Well, let's just use the first initial condition. y of 0 is

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equal to 2, which is equal to-- essentially just

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substitute 0 in into this equation.

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So it's c1 times e to the minus 2 times 0, that's

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essentially e to the 0, so that's just 1.

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So it's c1 times 1, which is c1, plus c2 times e to the

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minus 3 times 0.

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This is e to the 0, so it's just 1.

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So plus c2.

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So the first equation we get when we substitute our first

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initial condition is essentially c1 plus c2 is

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equal to 2.

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Now let's apply our second initial condition that tells

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us the slope at x is equal to 0.

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So y prime is 0.

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So this is our general solution, let's take its

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derivative, and then we can use this.

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So y prime of x is equal to what?

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The derivative of this is equal to minus 2 c1 times e to

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the minus 2x.

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And what's the derivative of this?

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It's minus 3 c2 times e to the minus 3x.

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And now we can use our initial condition, y prime at 0.

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So when x is equal to 0, what's the

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right-hand side equal?

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It's minus 2 times c1 and then e to the minus 0, e to the 0,

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that's just 1.

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Minus 3 c2, and then once again x is 0, so e to the

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minus 3 times 0, that's just 1.

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So it's just 1 times minus 3 c2.

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And it tells us that when x is equal to 0, what does this

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whole derivative equal?

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Well, it equals 3, right?

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Y prime of 0 is equal to 3.

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So now we go back into your first year of algebra.

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We have two equations-- two linear equations with two

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unknowns-- and we could solve.

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Let me write them in a form that you're

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probably more used to.

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So the first one is c1 plus c2 is equal to 2.

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And the second one is minus 2 c1 minus 3 c2 is equal to 3.

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So what can we do?

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Let's multiply this top equation by 2.

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There's a ton of ways to solve this, but if you multiply the

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top equation times 2, you'll get-- and I'll do this is a

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different color, just so that it's changed-- I'm just

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multiplying the top one by 2, you get 2 c1 plus 2 c2 is

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equal to 4.

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And now we can add these two equations.

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Minus 2 c1 plus 2, those cancel out.

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So minus 3 plus 2, you get minus c2 is equal to 7.

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Or we could say that c2 is equal to minus 7.

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And now we can substitute back in here.

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We have c1 plus c2-- c2 is minus 7-- so minus 7, is equal

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to 9, or we know that c-- oh sorry, no I'm already

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confusing myself.

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My brain was getting ahead of myself.

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c1 plus c2, that's minus 7, is equal to 2, right?

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I'm just substituting back into this differential

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equation-- sorry, to this equation, not a differential

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it's just a simple linear equation-- and then we get c1

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is equal to 9.

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And now we have our particular solution to the

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differential equation.

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So this was our general solution.

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We can just substitute our c1's and our c2's back in.

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We have our particular solution for those initial

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conditions.

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And I think that warrants a different color.

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So our particular solution is y of x is equal to c1, which

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we figured out is 9e to the minus 2x, plus c2-- well, c2

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is minus 7-- minus 7e to the minus 3x.

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That is the particular solution to our original

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differential equation.

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And it might be a good exercise for you to actually

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test it out.

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This particular solution to this differential equation.

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I'll do another example in the next video.

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I'll see you soon.

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