## anonymous 5 years ago d^2y/dx^2 - 2 dy/dx - 15y = 0 given y(0) = 10 y^1(0) = -7

1. anonymous

well the differential equation is: $y''-2y'-15y=0, y(0)=10, y'(0)=-7$ it can be easily solved using the auxiliary equations

2. anonymous

it's a 2nd order linear homogenous differential equation with constant co-efficents

3. anonymous

$m^2-2m-15=0 \implies (m-5)(m+3)=0 \implies m=5, m=-3$ the general solution is: $y(x)=c_1e ^{5x}+c_2e^{-3x}$ now we have just to solve for the initial conditions

4. anonymous

can you do the rest?

5. anonymous

I mean can you find c1 and c2? :)

6. anonymous

m^2 -2m -15 =0 (m+3)(m-5) gives you two real root solutions hence y=Ae^-3x + Be^5x no use the intial values to find A & B am i on the right path

7. anonymous

y= 57/8 e^-3x + 23/8 e^5x thats my answer

8. anonymous

$y'(x)=5c_1e ^{5x}-3c_2e ^{-3x} \implies y'(0)=5c_1-3c_2=10 \rightarrow (1)$ no need to continue... you got it already

9. anonymous

your answer is perfectly right.. you still have one thing to do, which is to be a fan ;)

10. anonymous

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