MAT244-2018S > Final Exam

FE-P3

(1/2) > >>

Victor Ivrii:
Find the general solution of
\begin{equation*}
y''' -6y'' +11y'- 6y=2\frac{e^{3x}}{e^x+1} .
\end{equation*}

Tim Mengzhe Geng:
First we find the solution for the homogeneous system

y^{(3)}-6y^{(2)}+11y^{(1)}-6y=0

The corresponding characteristic equation is

r^3-6r^2+11r-6=0

Three roots are

r_1=1

r_2=2

r_3=3

Then the solution for the homogeneous system is

y_c(t)=c_1e^{x}+c_2e^{2x}+c_3e^{3x}

where

y_1(t)=e^{x}

y_2(t)=e^{2x}

y_3(t)=e^{3x}

Then we follow to find the required solution to the nonhomogeneous equation. We use Variation of Parameters. We have

W[y_1,y_2,y_3]=2e^{6x}

W_1[y_1,y_2y_3]=e^{5x}

W_2[y_1,y_2y_3]=-2e^{4x}

W_3[y_1,y_2，y_3]=e^{3x}

and

g(x)=2\frac{e^{3x}}{e^{x}+1}

And then we have the following integration

\int \frac{W_1\cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{e^{2x}}{e^{x}+1}

\int\frac{e^{2x}}{e^{x}+1}=e^{x}-\ln(e^{x}+1)+c_4

\int \frac{W_2\cdot g(x)dx}{W[y_1,y_2,y_3]}=-2\int\frac{e^{x}}{e^{x}+1}

-2\int\frac{e^{x}}{e^{x}+1}=-2\ln(e^{x}+1)+c_5

\int \frac{W_3 \cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{1}{e^{x}+1}

\int\frac{1}{e^{x}+1}=x-\ln(e^{x}+1)+c_6

And finally, the required general solution $y(t)$

y(t)=\sum_{i=1}^3 y_i(t)\cdot\int\frac {W_i\cdot g(x)dx}{W[y_1,y_2,y_3]}

Meng Wu:
Small Error: $W_2(x)$ should be $2e^{4x}$.

Tim Mengzhe Geng:

--- Quote from: Meng Wu on April 11, 2018, 11:47:31 PM ---Small Error: $W_2(x)$ should be $2e^{4x}$.

--- End quote ---
For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$

Meng Wu:

--- Quote from: Tim Mengzhe GENG on April 11, 2018, 11:49:31 PM ---
--- Quote from: Meng Wu on April 11, 2018, 11:47:31 PM ---Small Error: $W_2(x)$ should be $2e^{4x}$.

--- End quote ---
For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$

--- End quote ---

Oh, you're right. My mistake.