Question

Solve \[y''+y'-2y=(3+4x)e^x\]
for \(y=y_c+y_p\)

by first finding \(y_c\) the complementry solution,
and then \(y_p\) the particular integral by;

i) using operator D

ii) using method of undertermined coeffecients

iii) using variation of parameters

iv) using the fact that \(y_1(x)=e^x\) satisfies the homgenouse equation

Answer 1

\[\begin{align*} y''+y'-2y &=(3+4x)e^x \\
\\
y_c''+y_c'-2y_c &=0 \\
m^2+m-2 &=0 \\
(m-1)(m+2) &=0 \\
\\
y_c=Ae^x+Be^{-2x}
\end{align*}\]

Answer 2

i)
\[\newcommand \D {\text D}
\begin{align*} y_p''+y_p'-2y_p &=(3+4x)e^x \\
\\
\text{Operator}\D \\
\\
(\D^2+\D-2)y_p &=(3+4x)e^x \\
y_p &=\frac1{\D^2+\D-2}(3+4x)e^x \\
&=e^x\frac1{(\D+1)^2+(\D+1)-2}(3+4x) \\
&=e^x\frac1{\D^2+2\D+1+\D+1-2}(3+4x) \\
&=e^x\frac1{\D^2+3\D}(3+4x) \\
&=e^x\frac1\D\frac1{\D+3}(3+4x) \\
&=\frac{e^x}3\frac1\D\frac1{1+\D/3}(3+4x) \\
&=\frac{e^x}3\frac1\D\Big(1-\D/3+\D^2/9-\dots\Big)(3+4x) \\%\D^3/27+
&=\frac{e^x}3\frac1\D\Big((3+4x)-4/3+0-\dots\Big) \\%0+
&=\frac{e^x}3\frac1\D\Big(\tfrac53+4x\Big) \\
&=\frac{e^x}3\int\tfrac53+4x\dd x \\
&=\frac{e^x}3\big(\tfrac53x+2x^2\big) \\%+c
y_p &=\big(\tfrac59x+\tfrac23x^2\big)e^x \\
\\
y(x) &=Ae^x+Be^{-2x}+\tfrac59xe^x+\tfrac23x^2e^x
\end{align*}\]

Answer 3

ii) \[\begin{align*} y_p''+y_p'-2y_p &=(3+4x)e^x \\
\\
\text{Method of undetermined coefficients} \\
\\
y_p =&(Dx+Ex^2)e^x \\
y'_p =&(Dx+Ex^2)e^x+(D+2Ex)e^x \\
y''_p =&(Dx+Ex^2)e^x+2(D+2Ex)e^x+(2E)e^x \\
\\
y_p''+y_p'-2y_p =&3(D+2Ex)e^x+(2E)e^x
=(3+4x)e^x \\
&\qquad(3D+2E)+(6E)x=3+4x \\
\\
&\qquad\qquad\quad3D+2E =3 \\
&\qquad\qquad\qquad\qquad\quad E =\tfrac46
=\tfrac23 \\
&\qquad\qquad\qquad D =\tfrac{3-4/3}{3}
=\tfrac59 \\
\\
y_p = &\left(\tfrac59x+\tfrac23x^2\right)e^x \\
\\
y = &Ae^x+Be^{-2x}+\tfrac59xe^x+\tfrac23x^2e^x
\end{align*}\]

Answer 4

iii)
\[\begin{align} y''+y'-2y &=(3+4x)e^x \\
\\
y_c''+y_c'-2y_c &=0 \\
m^2+m-2 &=0 \\
(m-1)(m+2) &=0 \\
\\
y_c=Ae^x+Be^{-2x}
\end{align}\]
\[\newcommand \dd [1] { \,\mathrm d#1 }
\begin{align} &\text{Variation of Parameters} \\
\\
W &= \begin{vmatrix} e^x & e^{-2x} \\
e^x & -2e^{-2x} \\
\end{vmatrix} =-2e^{-x}-e^{-x}
=-3e^{-x} \\
\\
y &=Ae^x+Be^{-2x} -e^x\int\frac{e^{-2x}(3+4x)e^x}{-3e^{-x}}\dd x
+e^{-2x}\int\frac{e^x(3+4x)e^x}{-3e^{-x}}\dd x \\
&=Ae^x+Be^{-2x} +\frac{e^x}3\int3+4x\dd x
-\frac{e^{-2x}}3\int(3+4x)e^{3x}\dd x \\
&=Ae^x+Be^{-2x}+\frac{e^x}3(3x+2x^2)
-\frac{e^{-2x}}3\Big(\frac{(3+4x)e^{3x}}{3}
-\int4\frac{e^{3x}}3\dd x\Big) \\
&=Ae^x+Be^{-2x}+(x+\tfrac23x^2)e^x
-\frac{e^{-2x}}3\Big((1+\tfrac43x)e^{3x}
-\tfrac49e^{3x}\Big) \\
&=Ae^x+Be^{-2x}+(x+\tfrac23x^2)e^x
-\tfrac13\Big(\tfrac59+\tfrac43x\Big)e^{x} \\
&=Ae^x+Be^{-2x}+(x+\tfrac23x^2)e^x
-\Big(\tfrac5{27}+\tfrac49x\Big)e^{x} \\
&=(A-\tfrac5{27})e^x+Be^{-2x}+(\tfrac59x+\tfrac23x^2)e^x \\
&=Ce^x+Be^{-2x}+\tfrac59xe^x+\tfrac23x^2e^x
\end{align}\]

Answer 5

iv)
\[\begin{align*} y''+y'-2y &=(3+4x)e^x \\
\\
\text{using}\qquad y_1 &=e^x \\
\\
y &=ue^x \\
y' &=\big(u'+u\big)e^x \\
y'' &=\big(u''+2u'+u\big)e^x \\
\\
y''+y'-2y &=\big(u''+3u'\big)e^x \\
&&\text{Let }u' &=p \\
&& u'' &=p' \\
y''+y'-2y &=\big(3p+p'\big)e^x
=(3+4x)e^x \\ p'+3p &=3+4x \\
&& R=e^{\int 3\dd x}=e^{3x} \\
(pe^{3x})' &=3e^{3x}+4xe^{3x} \\
pe^{3x} &=\int3e^{3x}+4xe^{3x}\dd x \\
pe^{3x} &=e^{3x}+\tfrac43xe^{3x}-\frac43\int e^{3x}\dd x \\
pe^{3x} &=e^{3x}+\tfrac43xe^{3x}-\tfrac49e^{3x}\\
pe^{3x} &=\tfrac59e^{3x}+\tfrac43xe^{3x}\\
p &=\tfrac59+\tfrac43x \\
u' &=\tfrac59+\tfrac43x \\
u &=\int\tfrac59+\tfrac43x\dd x \\
ye^{-x} &=\tfrac59x+\tfrac23x^2 \\
y &=\tfrac59xe^x+\tfrac23x^2e^x
\end{align*}\]

Answer 6

is there a mistake in method iii) Variation of Parameters with that constant term

Answer 7

should i have used y_p in iv)?

Answer 8

Doesn't variantion of paremeters involve useing the wronskian?

Answer 9

W=|...|

Answer 10

\(W_1 =\begin{bmatrix}
0 & y_1(x) & \\
f(x) & y'_1(x) &
\end{bmatrix}\) and \(W_2 =\begin{bmatrix}
y_2(x) & 0 & \\
y'_2(x) & f(x) &
\end{bmatrix}\)

to get \(u'_1 = \frac{W_1}{W}\) and \(u'_2 = \frac{W_2}{W}\) THEN INTEGRATE to get \(u_1\) and \(u_2\) or something ike that?

Answer 11

I think. Im a bit rusty. so, correct me if im wrong.

Answer 12

Im kinda new to variation of parameters but i was using

\[W(x)=\begin{vmatrix}y_1(x)&y_2(x)\\y'(x)&y_2'(x)\end{vmatrix}\]

where W is the wronskina of y_1 and y_2

Answer 13

and
\[y_p=-y_1(x)\int\frac{y_2(x)f(x)}{W(x)}\mathrm dx+y_2(x)\int\frac{y_1(x)f(x)}{W(x)}\mathrm dx\]

Answer 14

Ohhhh, essentually, you are doing the same thing, but I just dont like directly going into that formula, b/c my prof's didn't like using the formula, instead they wanted to see us use matrices to get to that.

Answer 15

so my wronskian is your two wronskians tangled together

Answer 16

Pretty much. \(W_1 =\begin{bmatrix}
0 & y_1(x) & \\
f(x) & y'_1(x) &
\end{bmatrix}\) = \(0(y'_(x))-y_1(f(x)) = u_1'\) but since you want \(u_1\) then, \(\int u_1'\) and then add \(u_2'\)

Answer 17

And I see you used the annihilator method for undetermined coefficients? Correct.