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# 2nd Order Linear Homogeneous Differential Equations 3: Let's use some initial conditions to solve for the particular solution

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- 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.

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