d^2y/dx^2 - 2 dy/dx - 15y = 0
given y(0) = 10 y^1(0) = -7

Question

d^2y/dx^2 - 2 dy/dx - 15y = 0
given y(0) = 10  y^1(0) = -7

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

can you do the rest?

anonymous · Answer

I mean can you find c1 and c2? :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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