Find the general solution to the following differential equation y''+3y′−4y=0

Question

Find the general solution to the following differential equation y''+3y′−4y=0

Answer 1

To find the general solution to the differential equation y'' + 3y' - 4y = 0, we'll use the characteristic equation method.

The characteristic equation for this differential equation is:

r^2 + 3r - 4 = 0

Now, we need to solve this quadratic equation for r. We can factor it as:

(r + 4)(r - 1) = 0

This gives us two distinct roots:

r1 = -4 r2 = 1

Since we have two distinct real roots, the general solution to the differential equation is a linear combination of the two exponential functions:

y(t) = C1 * e^(-4t) + C2 * e^(t)



where C1 and C2 are constants determined by initial conditions or boundary conditions.