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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?

\(\displaystyle r^2 -2r-3=0 .\)

\(\displaystyle y_h = c_1 e^{r_1 x} +c_2 e^{r_2 x}.\)

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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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.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?

\(\displaystyle r^2 -2r-3=0 .\)

\(\displaystyle y_h = c_1 e^{r_1 x} +c_2 e^{r_2 x}.\)

Could you help me with the particular solutions of the 4 exercises? I don't understand that yet.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.

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

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

\(\displaystyle y=y_h + y_p \) ; 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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Thank you very much. I already managed to solve exercise 1,3 and 4. The only one I cannot do is 2.1. 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 \(\displaystyle y_p =(ax+b)e^{2x}\).

\(\displaystyle [(ax+b)e^{2x}]''-2[(ax+b)e^{2x}]'-3(ax+b)e^{2x} =3xe^{2x}\) ; findaandb .

\(\displaystyle y=y_h + y_p \) ; 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 \(\displaystyle Ax^2 +Bx +C +D \sin(x) + E \cos(x)\).

Could you help me with that one?

\(\displaystyle 2(Ax^2 +Bx +C)''+3(Ax^2 +Bx+C)'+Ax^2 +Bx+C=Ax^2 +(6A+B)x+4A+C=x^2\) ; 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

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

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

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