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Find the general solution for x"+7x'+3x=0 with initial condition x(2)=1 ,x(2)=2

User Jospeh
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The general solution of the differential equation with the given initial conditions is x(t) = (1/2) * e^(-t) + (1/3) * e^(-3t).

Finding the general solution

Characteristic equation:

First, rewrite the differential equation in terms of the characteristic polynomial:

m^2 + 7m + 3 = 0

Factor the polynomial:

(m+1)(m+3) = 0

Therefore, the characteristic roots are m=-1, m=-3.

General solution:

The general solution for a homogeneous linear differential equation with constant coefficients is:

x(t) = c1 * e^(m1t) + c2 * e^(m2t)

where c1 and c2 are arbitrary constants and m1 and m2 are the characteristic roots.

Applying the characteristic roots:

Using the characteristic roots we found, the general solution becomes:

x(t) = c1 * e^(-t) + c2 * e^(-3t)

Finding the constants using initial conditions

Applying initial conditions:

We are given that x(2) = 1 and x'(2) = 2. Let's use these conditions to find the values of c1 and c2.

For x(2) = 1:

1 = c1 * e^(-2) + c2 * e^(-6)

For x'(2) = 2:

2 = -c1 * e^(-2) - 3c2 * e^(-6)

Solving the system of equations:

We can rewrite the system of equations as a matrix equation:

| e^(-2) e^(-6) | | c1 | | 1 |

| -e^(-2) -3e^(-6) | | c2 | | 2 |

Solving this system, we find:

c1 = 1/2 and c2 = 1/3

Final solution

Substituting the constants:

Substituting the values of c1 and c2 into the general solution, we get the final solution

x(t) = (1/2) * e^(-t) + (1/3) * e^(-3t)

Therefore, the general solution of the differential equation with the given initial conditions is:

x(t) = (1/2) * e^(-t) + (1/3) * e^(-3t)

User Udo Klein
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