# indeterminate coefficient method

#### Isken

Due to the quarantine we did not see this topic, the teacher let me solve these exercises and I have no idea how to do it, help please

#### idontknow

1. Find the characteristic roots of $$\displaystyle y'' _h -2y' _h -3y_h =0 ; \; \; y_h = ce^{rx}$$ , search the particular solution $$\displaystyle y_p = C e^{2x}$$.
the general solution is $$\displaystyle y=y_h +y_p$$. Can you continue?

$$\displaystyle r^2 -2r-3=0 .$$
$$\displaystyle y_h = c_1 e^{r_1 x} +c_2 e^{r_2 x}.$$

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topsquark and Isken

#### Isken

1. Find the characteristic roots of $$\displaystyle y'' _h -2y' _h -3y_h =0 ; \; \; y_h = ce^{rx}$$ , search the particular solution $$\displaystyle y_p = C e^{2x}$$.
the general solution is $$\displaystyle y=y_h +y_p$$. Can you continue?

$$\displaystyle r^2 -2r-3=0 .$$
$$\displaystyle y_h = c_1 e^{r_1 x} +c_2 e^{r_2 x}.$$
To be honest, no. Would it be annoying for you to help me with all the exercises? Excuse me if it's too much to ask.

#### skipjack

Forum Staff
1. Let's try $y = e^{2x}$. The left-hand side is $(2^2 - 2*2 - 3)e^{2x}$, i.e. $-3e^{2x}$, so $y = -e^{2x}$ is a partular solution. As $r^2 - 2r - 3 = (r + 1)(r - 3)$, the general solution is $y = \text{C}_1e^{-x} + \text{C}_2e^{3x} - e^{2x}$.

topsquark and Isken

#### Isken

1. Let's try y=e2xy=e2x. The left-hand side is (22−2∗2−3)e2x(22−2∗2−3)e2x, i.e. −3e2x−3e2x, so y=−e2xy=−e2x is a partular solution. As r2−2r−3=(r+1)(r−3)r2−2r−3=(r+1)(r−3), the general solution is y=C1e−x+C2e3x−e2xy=C1e−x+C2e3x−e2x.
Could you help me with the particular solutions of the 4 exercises? I don't understand that yet.

#### idontknow

1. search $$\displaystyle Y_p$$ as $$\displaystyle Ce^{2x}$$.
3. find the homogeneous solution $$\displaystyle y_h =c_1 e^{-x} +c_2 e^{3x}$$ then add it to particular solution.
Search the particular solution as $$\displaystyle y_p =(ax+b)e^{2x}$$.

$$\displaystyle [(ax+b)e^{2x}]''-2[(ax+b)e^{2x}]'-3(ax+b)e^{2x} =3xe^{2x}$$ ; find a and b .
$$\displaystyle y=y_h + y_p$$ ; take the derivative at point 0. $$\displaystyle y'(0)=0=y' _h (0) +y' _p (0)$$ and $$\displaystyle y(0)=1=y _h (0) +y _p (0)$$ to find the constants $$\displaystyle c_1 ,c_2$$.

4. search $Y_p$ as $$\displaystyle Ce^{-x} [A\sin (2x) + B \cos (2x) ]$$.

2. search $$\displaystyle Y_p$$ as $$\displaystyle Ax^2 +Bx +C +D \sin(x) + E \cos(x)$$.

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topsquark

#### Isken

1. search $$\displaystyle Y_p$$ as $$\displaystyle Ce^{2x}$$.
3. find the homogeneous solution $$\displaystyle y_h =c_1 e^{-x} +c_2 e^{3x}$$ then add it to particular solution.
Search the particular solution as $$\displaystyle y_p =(ax+b)e^{2x}$$.

$$\displaystyle [(ax+b)e^{2x}]''-2[(ax+b)e^{2x}]'-3(ax+b)e^{2x} =3xe^{2x}$$ ; find a and b .
$$\displaystyle y=y_h + y_p$$ ; take the derivative at point 0. $$\displaystyle y'(0)=0=y' _h (0) +y' _p (0)$$ and $$\displaystyle y(0)=1=y _h (0) +y _p (0)$$ to find the constants $$\displaystyle c_1 ,c_2$$.

4. search $Y_p$ as $$\displaystyle Ce^{-x} [A\sin (2x) + B \cos (2x) ]$$.

2. search $$\displaystyle Y_p$$ as $$\displaystyle Ax^2 +Bx +C +D \sin(x) + E \cos(x)$$.
Thank you very much. I already managed to solve exercise 1,3 and 4. The only one I cannot do is 2.
Could you help me with that one?

#### idontknow

Yp = polynomial-solution + trigonometric-solution .
$$\displaystyle 2(Ax^2 +Bx +C)''+3(Ax^2 +Bx+C)'+Ax^2 +Bx+C=Ax^2 +(6A+B)x+4A+C=x^2$$ ; equating the coefficients gives $$\displaystyle (A,B,C)=(1,-6,-4)$$.
$$\displaystyle y_1 =x^2 -6x -4$$ .
$$\displaystyle y_2$$ is the trigonometric solution . $$\displaystyle 2[ D \sin(x) + E \cos(x) ]''+3[ D \sin(x) + E \cos(x) ]'+ D \sin(x) + E \cos(x) =3\sin(x)$$ ; now find D and E .

topsquark

#### skipjack

Forum Staff
4. search $Y_p$ as $$\displaystyle Ce^{-x} [A\sin (2x) + B \cos (2x) ]$$.
That doesn't suffice.

idontknow

#### idontknow

That doesn't suffice.
can you corrrect it ? so that the OP is not mistaken .

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