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Solve IVP 2y"'-6y" +7y' 3y = 0, y(0) = 1, y'(0) = 0, y" (0) = -5

Sagot :

Thakurayushanand3idn

Answer:

Step-by-step explanation:

This is a second degree, homogeneous differential Equation.

Here’s how this works:

Set y = e^(rx) .

Then, y’ = r*e^(rx) and y’’ = r^2*(e^(rx)).

When I plug everything in, I get the following characteristic equation:

r^2 +3*r +7 = 0

Solving for r, I get

r1 = (-3 +i*sqrt(19))/2 and r2 = (-3-i*sqrt(19))/2

So, your solution will look like the following:

y = e^(-3/2*x) [A*cos(x*sqrt(19)/2) + B*sin(x*sqrt(19)/2)]

Now, you want to use the initial conditions to solve for A and B.

y(0) = 1 means when x =0, y = 1.

We know that A = 1.

y’(0) = 2 means when x = 0, y’ = 2.

To solve for B, you can take the derivative. It involves the product Rule. Replace x with 0 and y’ with 2. You’ll be able to find B. I hope this helps.

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