Answer: The required solution of the given IVP is

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

Let
be an auxiliary solution of the given differential equation.
Then, we have

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}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmx%7D-e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2-1%29e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2-1%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmx%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%3D1%5C%5C%5C%5C%5CRightarrow%20m%3D%5Cpm1.)
So, the general solution of the given equation is
where A and B are constants.
This gives, after differentiating with respect to x that

The given conditions implies that

and

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

From equation (i), we get

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

Thus, the required solution of the given IVP is
