Find the general solution of y"+4y'-2y=0
To find the general solution of a second-order linear homogeneous differential equation like y"+4y'-2y=0, we can follow these steps:
Step 1: Write the characteristic equation.
The characteristic equation is obtained by assuming a solution of the form y = e^(rt), where r is an arbitrary constant.
For the given equation, the characteristic equation is r^2 + 4r - 2 = 0.
Step 2: Solve the characteristic equation.
To solve the characteristic equation, we can use the quadratic formula or factorization.
Using the quadratic formula, r = (-b ± √(b^2 - 4ac))/(2a), where a = 1, b = 4, and c = -2.
Thus, r = (-4 ± √(16 + 8))/2 = (-4 ± √24)/2 = -2 ± √6.
So, the roots of the characteristic equation are r1 = -2 + √6 and r2 = -2 - √6.
Step 3: Write the general solution.
Since the characteristic equation has distinct real roots, the general solution is given by y = C1e^(r1t) + C2e^(r2t), where C1 and C2 are constants to be determined.
Therefore, the general solution to the given differential equation is y = C1e^((-2 + √6)t) + C2e^((-2 - √6)t), where C1 and C2 are arbitrary constants.