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Find the value?

Consider the initial value problem: y'+(2/3)y=1-(1/2)t, y(0)=yο. Find the value of yο for which the solution touches, but does not cross, the t-axis.

Comments

This is a very challenge problem. Besides algebra, you need to have y' = 0 at y = 0. You can graph my solution and check the answer.

 

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1 Answer

Use the solution formula for the first order linear differential equation,

y = e^(2t/3)[∫e^(-2t/3)(1-(1/2)t) dt + c] = (3/8)(2t-1) + ce^(2t/3), integration by parts

Plug in y(0) = yo,

yo = -3/8 + c => c = yo+3/8

y = (3/8)(2t-1) + (y0+3/8)e^(2t/3)

y' = 0 => 3/4 + (y0+3/8)(2/3)e^(2t/3) = 0 ......(1)

(3/8)(2t-1) + (y0+3/8)e^(2t/3) = 0 ......(2)

Solve the system of (1) and (2),

(3/8)(2t-1) = 9/8

t = 2

Plug into (2),

(3/8)(3) + (y0+3/8)e^(4/3) = 0

Solve for yo,

yo = -(9/8)e^(-4/3) - 3/8  <==Answer