# b.1.3.7 Solve IVP

#### karush

##### Well-known member
$\tiny{b.1.3.7}$
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...

#### topsquark

##### Well-known member
MHB Math Helper
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

##### Well-known member
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

##### Well-known member
MHB Math Helper
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

#### Country Boy

##### Well-known member
MHB Math Helper
$Ae^{1- t}= Ae^1e^{-t}$ 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'

MHB Math Helper
@karush: