Question

Solve y'''+2y''-y'-2y=e^2x then y=​

Answer 1

Answer:

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Answer 2

Answer:

Find the general solution of the second-order differential equation

 \[ y'' + y' - 2y = e^{2x}. \]

If the solution is not valid everywhere, describe the interval on which it is valid.

The general solution of the homogeneous equation

 \[ y'' + y' - 2y = 0 \]

is given by Theorem 8.7 with a = 1 and b = -2. This gives us d = a^2 - 4b = 9; hence, k = \frac{1}{2} \sqrt{d} = \frac{3}{2}. Thus,

 \begin{align*}  y &= e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx}) \\  &= e^{-\frac{x}{2}} ( c_1 e^{\frac{3}{2}x} + c_2 e^{-\frac{3}{2}x}) \\  &= c_1 e^x + c_2 e^{-2x}. \end{align*}

To find a particular solution of y'' + y' - 2y = e^{2x} assume y_1 = p(x)e^{2x} is a solution. Then,

 \begin{align*}  y_1' &= 2p(x) e^{2x} + p'(x)e^{2x} \\  y_1'' &= p''(x) e^{2x} + 4p'(x) e^{2x} + 4p(x) e^{2x}. \end{align*}

Therefore,

 \begin{align*}  &&p''(x)e^{2x} + 4p'(x) e^{2x} + 4p(x) e^{2x} + 2p(x) e^{2x} + p'(x)e^{2x} - 2p(x)e^{2x} &= e^{2x} \\  \implies && 4p(x) + 5p'(x) + p''(x) &= 1.  \end{align*}

Then, p(x) = \frac{1}{4} is as solution to this so we have

 \[ y_1 = \frac{1}{4} e^{2x}. \]

Hence, the general solution is given by

 \[ y = c_1 e^x + c_2 e^{-2x} + \frac{1}{4} e^{2x}. \]

Step-by-step explanation: