Question

Solve 5y′′−12y′+4y=0,y(0)=5,y′(0)=6.8
y(t)=?

Answer 1

To solve the given second-order differential equation with initial conditions, we find the characteristic equation's roots, construct the general solution, and then use the initial conditions to find the particular solution.

Step-by-step explanation:

The student has presented a second-order linear homogeneous differential equation with initial conditions: 5y'' - 12y' + 4y = 0, y(0) = 5, y'(0) = 6.8. To solve this differential equation and find the function y(t), we perform the following steps:

1. Write the characteristic equation which is 5r^2 - 12r + 4 = 0.
2. Find the roots of the characteristic equation, which will be the values of r that satisfy the equation.
3. With the roots, construct the general solution to the differential equation. The form of the solution depends on whether the roots are real and distinct, a real repeated root, or complex conjugate pairs.
4. Apply the initial conditions to solve for any constants in the general solution.
5. Write the particular solution that satisfies the initial conditions.

In this case, the characteristic equation yields two distinct real roots, allowing us to write the solution as y(t) = C1*e^(r1*t) + C2*e^(r2*t), where C1 and C2 are constants determined by the initial conditions.