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Find the solution of the given initial value problem:

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

User Korbi
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1 Answer

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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).

User MarceloBarbosa
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