Consider the following system of two first-order ODEs for y1(t) and y2(t) , y′1y′2=y1+8y2=2y1+3y2. Where y(t)=y1 , this system of differential equation is equivalent to a second-order ODE for y(t) of the form y′′−4y′+cy=0. What is the value of c ? Enter your integer valued answer below.

时间: 2023-06-26 18:05:37 浏览: 148
We can start by solving for y2 in the first equation: y′1y′2 = 2y1 + 3y2 y′2 = (2y1 + 3y2) / y′1 Substituting this expression for y′2 into the second equation: y′1(2y1 + 3y2) / y′1 = y1 + 8y2 2y1 + 3y2 = y1 + 8y2 y1 = 5y2 Substituting this relationship back into the first equation: y′1y′2 = 2y1 + 3y2 y′1y′2 = 2(5y2) + 3y2 y′1y′2 = 13y2 Now we can differentiate both sides with respect to t: y′′1y′2 + y′1y′′2 = 13y′2 Substituting y′2 = (2y1 + 3y2) / y′1: y′′1(2y1 + 3y2) / y′1 + y′1(3y′1(2y1 + 3y2) - (2y1 + 3y2)y′′1) / y′1^2 = 13(2y1 + 3y2) / y′1 Simplifying: 2y′′1 + 9y′1^2 / y′1 - 3y′′1 = (26y1 + 39y2) / y′1 y′′1 - (4y′1) + (13/5)y1 = 0 Comparing with the standard form of a second-order ODE: y′′ + py′ + qy = 0 We can see that c = 13/5.
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