Question

Find the solution of the given initial value problem:

y''- y = 0, y(0) = 2, y'(0) = -1/2

Answer 1

Answer: The required solution of the given IVP is

`y(x)=(3)/(4)e^x+(5)/(4)e^(-x).`

Step-by-step explanation: We are given to find the solution of the following initial value problem :

`y^(\prime\prime)-y=0,~~~y(0)=2,~~y^\prime(0)=-(1)/(2).`

Let
`y=e^(mx)` be an auxiliary solution of the given differential equation.

Then, we have

`y^\prime=me^(mx),~~~~~y^(\prime\prime)=m^2e^(mx).`

Substituting these values in the given differential equation, we have

`m^2e^(mx)-e^(mx)=0\\\\\Rightarrow (m^2-1)e^(mx)=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^(mx)\\eq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.`

So, the general solution of the given equation is

`y(x)=Ae^x+Be^(-x),` where A and B are constants.

This gives, after differentiating with respect to x that

`y^\prime(x)=Ae^x-Be^(-x).`

The given conditions implies that

`y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)`

and

`y^\prime(0)=-(1)/(2)\\\\\\\Rightarrow A-B=-(1)/(2)~~~~~~~~~~~~~~~~~~~~~~~~(ii)`

Adding equations (i) and (ii), we get

`2A=2-(1)/(2)\\\\\\\Rightarrow 2A=(3)/(2)\\\\\\\Rightarrow A=(3)/(4).`

From equation (i), we get

`(3)/(4)+B=2\\\\\\\Rightarrow B=2-(3)/(4)\\\\\\\Rightarrow B=(5)/(4).`

Substituting the values of A and B in the general solution, we get

`y(x)=(3)/(4)e^x+(5)/(4)e^(-x).`

Thus, the required solution of the given IVP is

`y(x)=(3)/(4)e^x+(5)/(4)e^(-x).`