Solve this second order differential equation, with conditions ?

Question

d^2y/dx^2 + 4dy/dx + 3y = 0

subject to the conditions y(0)= 0 and y'(0)=1

Answer 1

The solution is of form y=ke^ax (a may be complex)

dy/dx = ae^ax

dy^2/dx^2 = a^2 e^ax

a^2 e^ax + 4ae^ex + 3e^ax = 0

a^2 + 4a + 3 = 0

a^2 + 4a + 4 - 4 + 3 = 0

(a + 2)^2 - 1 = 0

(a + 1)(a + 3) = 0

Now, the solution is:

y = ke^-x + me^-3x

y(0) = 0 --> k+m = 0

y'(x) = -ke^-x -3me^-3x  --> y'(0) = -k - 3m

Let's solve the system

k + m = 0

-k - 3m = 1

Adding them:

-2m = 1

m = -1/2

From k + m = 0, we get that k=1/2

y = (e^-x)/2 - (e^-3x)/2

Answer 2

d^2y/dx^2(0)=-4dy/dx(0)-3y(0)

d^2y/dx^2(0)=-4*1-3*0

d^2y/dx^2(0)=-4

Answer 3

y'' + 4y' + 3y = 0

You need solutions to this form of differential equation such that the second derivative has the same form as the first such that it can cancel to zero.

What function has a derivative that is the same as the function itself? The exponential function.

So solutions are in the form of e^(ct) where c is a constant to be determined.

In taking the derivative, you will bring down c in the first and c^2 in the second. Thus, by factoring out the c terms and noting that the e terms cannot equal zero, you will get a quadratic equal to zero called the "characteristic equation."

So multiplying the powers of c by the given constants.

c^2 + 4c + 3 = 0

Factoring

(c+3)(c+1) = 0

c= -1,-3

So you have the 2 possible solutions. Since this is a linear differential equation, you can superimpose the individual solutions to find the total solution (y). If each one is individually a solution, then plugging them into the equation will give 0 + 0 = 0, preserving the equality and showing that superimposed solutions are also solutions.

So y(t) = c1 * y1 + c2*y2 = c1*e^(-t) + c2*e^(-3t)

Use the initial conditions to solve:

So y(0) = 0 --> c1 + c2 = 0

y'(t) = -c1*e^(-t) - 3c2*e^(-3t)

So y'(0) = 1 --> -c1 - 3c2 = 1

So from the first, c1 = -c2

Substituting:

c2-3c2 = 1

-2c2 = 1

c2 = -1/2

And c1 thus equals 1/2.

So y(t) = 1/2*e^(-t) -1/2*e^(-3t).