Question

Really struggling on how to determine the particular solution of the right hand side of the equation. How do you know where to start when solving for the Yp(t) equation?

5. Determine a real-valued fundamental set of solutions and the general solution of the nonhomogenous differential equation y"(t) – 2y'(t) + 10y(t) = 20² + 2t – 8

Answer 1

y" – 2y + 10y = 20t2 + 2t – 8

for homogeneous system find roots

2 - 2.r + 10 = 0

-6+ vb? – 4ac r1. 2 2a

- (-2) + V(-2)? – 4.1. 10 2.1 r1, 2 =

2+ V4 - 40 r1,2 2.

$x_{1,\:2}=\frac{2\pm \:\sqrt{-36}}{2}$

$x_{1,\:2}=\frac{2\pm 6i}{2}$

$x_{1,\:2}=1\pm 3i$

for complex roots complementary solution is

${\color{Red} y_c=e^t\left(c_1\cos \left(3t\right)+c_2\sin \left(3t\right)\right)}$

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on the right side we have $g\left(t\right)=20t^2+2t-8$

so we assume that a particular solution is

$y=a_0t^2+a_1t+a_2$........................(1)

take the first derivative

take the second derivative

$y''=2a_0$

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put all values in the given DE

y" – 2y + 10y = 20t2 + 2t – 8

$2a_0-4a_0t-2a_1+10a_0t^2+10a_1t+10a_2=20t^2+2t-8$

$10a_0t^2+\left(-4a_0+10a_1\right)t+\left(2a_0-2a_1+10a_2\right)=20t^2+2t-8$

compare coefficient both sides

$20=10a_0$

${\color{Blue} a_0=2}$

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$2=-4a_0+10a_1$

$2=-4 \cdot 2+10a_1$

$2+8=10a_1$

$10=10a_1$

${\color{Blue} a_1=1}$

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$-8=2a_0-2a_1+10a_2$

$-8=2 \cdot 2-2\cdot 1+10a_2$

${\color{Blue} a_2=-1}$

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put all constant in equation 1

$y=a_0t^2+a_1t+a_2$

$y=2t^2+1\cdot \:t-1$

${\color{Red} y_p= 2t^2+t-1}$

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general solution is

$y=y_c+y_p$

${\color{Red} y= e^t\left(c_1\cos \left(3t\right)+c_2\sin \left(3t\right)\right)+2t^2+t-1}$