Answer on Question #69648 – Math – Differential Equations
Question
1. Solve the differential equation
2dx2d2y+5dxdy−12y=0
Solution
2y′′+5y′−12y=0
1. Construct a characteristic equation:
2k2+5k−12=0;k1=−4;k2=23;
2. The general solution of the differential equation:
y=C1e−4x+C2e23x,
where C1 and C2 are arbitrary real constants.
Answer: y=C1e−4x+C2e23x.
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