Use the Laplace transform to solve the following initial value problem: y"-4y'+5y=0 y(0)=1, y'(0)=2
Using Y for the Laplace transform of y(t), i. e. Y=L{y(t)},
find the equation you get by taking the Laplace transform of the differential equation
=0
Now solve for Y(s)=
By completing the square in the denominator and inverting the transform, find y(t)=

Laplace transform of the original differential equation:

.

Given that and , solve for to obtain:

.

Apply inverse Laplace transform to obtain:

.

Step-by-step explanation:

Apply these two rules to replace all , , and in the original equation with their Laplace transforms:

.

Substitute in the values and .

.

Solve for after rearranging this equation:

.

Note that if denominator is the left-hand side of a quadratic equation, this equation would have no real root. Hence, complete the square in the denominator:

.

Look up a table of Laplace transforms. Apply the rule , where and .

.

answer: ouch. thats difficult

step-by-step explanation: