Solving IVP $y''-y=0$ with $y_1,y_2$

[karush]

$\tiny{b.1.3.7}$
Solve IVP
$y''-y=0;\quad y_1(t)=e^t,\quad y_2(t)=\cosh{t}$
$\begin{array}{lll}
&\exp\left(\int \, dx\right)= e^x\\
& e^x(y''-y)=0\\
& e^x-e^x=0\\ \\
&y_1(x)=e^x\\
&(e^x)''-(e^x)=0\\
&(e^x)-(e^x)=0\\ \\

&y_2(x)=\cosh{x}\\
&(\cosh{x})''-(\cosh{x})=0\\
\end{array}$

ok there was no book answer so hopefully went in right direction so suggestions...:unsure:

 

[topsquark]

That method doesn't work for second degree equations. Use the method of characteristics.

We have y'' - y = 0. So write \(\displaystyle m^2 - 1 = 0\). This has solutions m = -1, 1. Thus the homogeneous solution will be
\(\displaystyle y_h = A e^{1 \cdot t} + B e^{-1 \cdot t}\)

To fit the particular solution you can look at the method of undetermined coefficients. Hint: Let \(\displaystyle cosh(t) = \dfrac{1}{2} ( e^{t} + e^{-t} )\).

-Dan

 

[karush]

That method doesn't work for second degree equations. Use the method of characteristics.

We have y'' - y = 0. So write \(\displaystyle m^2 - 1 = 0\). This has solutions m = -1, 1. Thus the homogeneous solution will be
\(\displaystyle y_h = A e^{1 \cdot t} + B e^{-1 \cdot t}\)

To fit the particular solution you can look at the method of undetermined coefficients. Hint: Let \(\displaystyle cosh(t) = \dfrac{1}{2} ( e^{t} + e^{-t} )\).

-Dan

so we can do this?
$\dfrac{1}{2} ( e^{t} + e^{-t} )=A e^{1 \cdot t} + B e^{-1 \cdot t}$

 

[topsquark]

Hmmm... Apparently the whole method isn't contained in what I quoted.

If you have a term \(\displaystyle Ae^{t}\) in \(\displaystyle y_h\) then you can't have that term in \(\displaystyle y_p\). (Think about it... If you use \(\displaystyle e^t\) in your particular solution then it will just give 0 because it is already in the homogeneous solution.) So instead of using \(\displaystyle e^t\) try \(\displaystyle te^t\). If that doesn't work or if you need it (in this case you will) try \(\displaystyle t^2 e^t\), etc.

-Dan

 

[HOI]

$Ae^{1- t}= Ae^1e^{-t}$ and the constant, e, can be "absorbed" into "A"; $A'= Ae$ so $Ae^{1- t}= A'e^{-t}$.
Similarly $Be^{1- t}= B'e^{-t}$ with $B'= Be$.

And, of course those can be combined: $A'e^{-t}+ B'e^{-t}= Ce^{-t}$ with C= A'+ B'

 

[topsquark]

@karush:
What Country Boy said. I misread your problem.

-Dan