5y/4y-2/2 -10y/8y-1/5=0
Answers
Step-by-step explanation:
Find the general solution of the second-order differential equation
\[ y'' - 5y' + 4y = x^2 - 2x + 1. \]
If the solution is not valid everywhere, describe the interval on which it is valid.
The general solution of the homogeneous equation
\[ y'' - 5y' + 4y = 0 \]
is given by Theorem 8.7 with a = -5 and b = 4. 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{5}{2}x} (c_1 e^{\frac{3}{2}x} + c_2 e^{-\frac{3}{2} x} ) \\ &= c_1 e^{4x} + c_2 e^x. \end{align*}
To find a particular solution of y'' - 5y' + 4y = x^2 - 2x + 1 let y_1 = Ax^2 + Bx + C. Then,
\begin{align*} y_1' &= 2Ax + B \\ y_1'' &= 2A. \end{align*}
Therefore,
\[ y'' - 4y' + 5y = x^2 - 2x + 1 \quad \implies \quad 2A - 10Ax - 5B + 4Ax^2 + 4Bx + 4C = x^2 - 2x + 1. \]
Answer:
5y-2/4y+2-10y-1/8y-5=0