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

• MHB
• karush
In summary, the conversation revolves around solving the initial value problem (IVP) of a second degree equation with variables $y''-y=0$. The suggested method is to use the method of characteristics and determining the homogeneous solution, which is given by $y_h = Ae^{t} + Be^{-t}$, where $A$ and $B$ are constants. To find the particular solution, the method of undetermined coefficients is suggested, with a hint to use $\cosh{t} = \dfrac{1}{2} ( e^{t} + e^{-t} )$ as a possible solution. However, it is pointed out that this term cannot be used in the particular solution if it is already
karush
Gold Member
MHB
$\tiny{b.1.3.7}$
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:

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

topsquark said:
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?
$\dfrac{1}{2} ( e^{t} + e^{-t} )=A e^{1 \cdot t} + B e^{-1 \cdot t}$

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

$Ae^{1- t}= Ae^1e^{-t}$ 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'

@karush:

-Dan

## 1. What is an IVP?

An IVP, or initial value problem, is a type of differential equation that involves finding a solution to an equation with an initial condition. In other words, we are given a function and its derivative at a specific point, and we need to find a function that satisfies the equation and also satisfies the initial condition.

## 2. What is the general process for solving an IVP?

The general process for solving an IVP involves using techniques such as separation of variables, integrating factors, or power series to find a general solution to the equation. Then, we use the initial condition to determine the specific solution that satisfies the equation and the initial condition.

## 3. How do we solve an IVP with $y''-y=0$ and $y_1, y_2$?

For this specific IVP, we can use the characteristic equation method. This involves finding the roots of the characteristic equation $r^2-1=0$, which are $r=1$ and $r=-1$. Then, our general solution will be of the form $y=c_1e^x+c_2e^{-x}$. We can then use the initial conditions $y(0)=y_1$ and $y'(0)=y_2$ to determine the specific solution.

## 4. Can we solve an IVP with $y''-y=0$ analytically?

Yes, we can solve this IVP analytically using the method described in the previous question. However, for more complicated IVPs, it may not always be possible to find an analytical solution and numerical methods may be used instead.

## 5. What are some real-world applications of solving IVPs?

IVPs are used in many fields of science and engineering, such as physics, biology, and economics. For example, in physics, IVPs are used to model the motion of objects under the influence of forces. In biology, IVPs are used to model population growth or the spread of diseases. In economics, IVPs are used to model economic growth or the effects of policies on a market.

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