indeterminate coefficient method

Mar 2020
4
0
Mexico
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
received_201766207795471.png
 
Dec 2015
1,068
164
Earth
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?

\(\displaystyle r^2 -2r-3=0 .\)
\(\displaystyle y_h = c_1 e^{r_1 x} +c_2 e^{r_2 x}.\)
 
Last edited:
  • Like
Reactions: topsquark and Isken
Mar 2020
4
0
Mexico
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
Dec 2006
21,468
2,463
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 $y = \text{C}_1e^{-x} + \text{C}_2e^{3x} - e^{2x}$.
 
  • Like
Reactions: topsquark and Isken
Mar 2020
4
0
Mexico
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.
 
Dec 2015
1,068
164
Earth
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 \(\displaystyle Ax^2 +Bx +C +D \sin(x) + E \cos(x)\).
 
Last edited:
  • Like
Reactions: topsquark
Mar 2020
4
0
Mexico
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?
 
Dec 2015
1,068
164
Earth
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 .
 
  • Like
Reactions: topsquark