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Find the solution of with y(0) = 2 and y' (0) = 3. y= y"-6y +9y = 150 et

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User Arelius
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r^2 - 6r + 9 - 150 / e^(rt) = 0 is the solution . We need to find the solution of this second-order linear homogeneous differential equation with the initial conditions y(0) = 2 and y'(0) = 3.

Taking the derivatives of y, we have y' = re^(rt) and y" = r^2e^(rt).

Substituting these derivatives into the differential equation, we get:

r^2e^(rt) - 6re^(rt) + 9e^(rt) = 150.

Factoring out e^(rt), we have:

e^(rt)(r^2 - 6r + 9) = 150.

Since e^(rt) is never equal to zero, we can divide both sides of the equation by e^(rt):

r^2 - 6r + 9 = 150 / e^(rt).

Simplifying further, we have:

r^2 - 6r + 9 - 150 / e^(rt) = 0.

This is a quadratic equation in terms of r. Solving for r using the quadratic formula, we find two possible values for r.

To learn more about differential equation click here :

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User Neemzy
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