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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well the differential equation is: \[y''2y'15y=0, y(0)=10, y'(0)=7\] it can be easily solved using the auxiliary equations

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's a 2nd order linear homogenous differential equation with constant coefficents

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[m^22m15=0 \implies (m5)(m+3)=0 \implies m=5, m=3\] the general solution is: \[y(x)=c_1e ^{5x}+c_2e^{3x}\] now we have just to solve for the initial conditions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean can you find c1 and c2? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0m^2 2m 15 =0 (m+3)(m5) 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
 5 years ago
Best ResponseYou've already chosen the best response.0y= 57/8 e^3x + 23/8 e^5x thats my answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y'(x)=5c_1e ^{5x}3c_2e ^{3x} \implies y'(0)=5c_13c_2=10 \rightarrow (1)\] no need to continue... you got it already

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your answer is perfectly right.. you still have one thing to do, which is to be a fan ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try www.aceyourcollegeclasses.com
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