Uploaded by TheIntegralCALC on 19.02.2011

Transcript:

Hi everyone. Welcome back to integralcalc.com. Today we’re going to be doing a second-order

differential equations problem. The function that we're given is y'' - 7y' + 10y = 0.

And as with any second-order differential equation problem, the formula that we’re

going to be using is the following. y(x) = {c_1}e^{(r_1)x} + {c_2}e^{(r_2)x}

The way that we're going to use this formula, our answer will still contain c sub 1 and

c sub 2. What we're looking to solve for is r_1 and r_2. We want to replace those with

actual values from our problem. And the way that we do that is by transforming our original

equation here into something that we can factor. So we replace the y's with r's.

y -----> r When we do that, we have just y on its own

here; we're actually going to treat it as 1.

y’’ - 7y’ + 10y = 0 y’’ - 7y’ + 10(1) = 0

y’’ - 7y’ + 10 = 0 And then y’, we'll treat as r.

y’’ -7r + 10 = 0 y” we'll treat as r^2 which on the one hand

is a little confusing. y’’ - 7r + 10 = 0

{r^2} - 7r + 10 = 0 If it's tough for you to remember,I think

the easiest trick is notice that with the y here (10y), you have no little prime mark,

it's just the original y so there's 0 little dashes on it. It’s a ridiculous way to remember

it but y to the 0 is 0y. Here (7y’) we have one dash so y to the 1 is just y which we

replace with r. Here (y’’) we have two dashes, y to the 2 is y squared we replace

with r, we get r^2. So basically, you're just kind of counting

the number of dashes and y” gets replaced with r^2, y’ gets replaced with r and y

gets replaced with 1. We make each of those replacements and then

we factor our function, in this case the factoring is really easy.

(r-5) (r-2) = 0 We end up with r minus 5 times r minus 2 and

when we solve for r, we get 2 solutions. One is 5 and one is 2.

r = 5, 2 You’re always going to get two solutions

when you do a second-order differential equations problem and you're going to plug one in for

r_1 and the other in for r_2. Remember we're going to leave c_1 and c_2, but we end up

replacing r_1 and r_2 with the solutions that we just found so we end up with c_1 e^5x +

c_2 e^2x. y(x) = {c_1}e^{(r_1)x} + {c_2}e^{(r_2)x}

r = 5, 2 y(x) = {c_1}e^{5x} + {c_2}e^{2x}

And that's our final answer. So I hope that helped you, guys and I'll see

you in the next problem. Bye.

differential equations problem. The function that we're given is y'' - 7y' + 10y = 0.

And as with any second-order differential equation problem, the formula that we’re

going to be using is the following. y(x) = {c_1}e^{(r_1)x} + {c_2}e^{(r_2)x}

The way that we're going to use this formula, our answer will still contain c sub 1 and

c sub 2. What we're looking to solve for is r_1 and r_2. We want to replace those with

actual values from our problem. And the way that we do that is by transforming our original

equation here into something that we can factor. So we replace the y's with r's.

y -----> r When we do that, we have just y on its own

here; we're actually going to treat it as 1.

y’’ - 7y’ + 10y = 0 y’’ - 7y’ + 10(1) = 0

y’’ - 7y’ + 10 = 0 And then y’, we'll treat as r.

y’’ -7r + 10 = 0 y” we'll treat as r^2 which on the one hand

is a little confusing. y’’ - 7r + 10 = 0

{r^2} - 7r + 10 = 0 If it's tough for you to remember,I think

the easiest trick is notice that with the y here (10y), you have no little prime mark,

it's just the original y so there's 0 little dashes on it. It’s a ridiculous way to remember

it but y to the 0 is 0y. Here (7y’) we have one dash so y to the 1 is just y which we

replace with r. Here (y’’) we have two dashes, y to the 2 is y squared we replace

with r, we get r^2. So basically, you're just kind of counting

the number of dashes and y” gets replaced with r^2, y’ gets replaced with r and y

gets replaced with 1. We make each of those replacements and then

we factor our function, in this case the factoring is really easy.

(r-5) (r-2) = 0 We end up with r minus 5 times r minus 2 and

when we solve for r, we get 2 solutions. One is 5 and one is 2.

r = 5, 2 You’re always going to get two solutions

when you do a second-order differential equations problem and you're going to plug one in for

r_1 and the other in for r_2. Remember we're going to leave c_1 and c_2, but we end up

replacing r_1 and r_2 with the solutions that we just found so we end up with c_1 e^5x +

c_2 e^2x. y(x) = {c_1}e^{(r_1)x} + {c_2}e^{(r_2)x}

r = 5, 2 y(x) = {c_1}e^{5x} + {c_2}e^{2x}

And that's our final answer. So I hope that helped you, guys and I'll see

you in the next problem. Bye.