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Solve:
[ yʹ'-4 yʹ+4 y=0, y(0)=1, yʹ(0)=1 .; y(1/3)=? ]

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Final answer:

To solve the given differential equation y'' - 4y' + 4y = 0 with initial conditions, we use the characteristic equation method and find the general solution to be y = c1 * e^(2x) + c2 * x * e^(2x). Substituting the initial conditions, we find c1 = 1 and c2 = 0, giving us the specific solution y = e^(2x). Finally, substituting x = 1/3, we find y(1/3) ≈ 2.71828.

Step-by-step explanation:

To solve the given differential equation y'' - 4y' + 4y = 0, with initial conditions y(0) = 1 and y'(0) = 1, we can use the characteristic equation method. The characteristic equation for this equation is r^2 - 4r + 4 = 0. Solving this quadratic equation, we find that r = 2.

Therefore, the general solution of the differential equation is y = c1 * e^(2x) + c2 * x * e^(2x). Using the initial conditions, we can find the values of c1 and c2. Substituting x = 0, y = 1, and x = 0, y' = 1, into the general solution, we get the following equations: c1 + c2 * 0 = 1 and 2c1 + c2 = 1.

Solving these equations, we find that c1 = 1 and c2 = 0. Therefore, the specific solution for the given differential equation is y = e^(2x). Substituting x = 1/3 into this equation, we find that y(1/3) = e^(2/3) = approximately 2.71828.

User Francis Nepomuceno
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