Question

Find the general solution to y'" - y" + 2y = 0

Answer 1

Answer: The required general solution is

`y(x)=Ae^(-x)+e^x(B\cos x+C\sin x).`

Step-by-step explanation: We are given to find the general solution of the following differential equation :

`y^(\prime\prime\prime)-y^(\prime\prime)+2y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)`

Let y = y(x) and
`y=e^(mx)` be an auxiliary solution of equation (i).

Then, we have

`y^\prime=me^(mx),~~~y^(\prime\prime)=m^2e^(mx),~~~y^(\prime\prime\prime)=m^3e^(mx).`

Substituting these values in equation (i), we have

`m^3e^(mx)-m^2e^(mx)+2e^(mx)=0\\\\\Rightarrow (m^3-m^2+2)e^(mx)=0\\\\\Rightarrow m^3-m^2+2=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^(mx)\\eq0]\\\\\Rightarrow m^2(m+1)-2m(m+1)+2(m+1)=0\\\\\Rightarrow (m+1)(m^2-2m+2)=0\\\\\Rightarrow m+1=0~~~~~\Rightarrow m=-1`

and

`m^2-2m+2=0\\\\\Rightarrow (m^2-2m+1)+1=0\\\\\Rightarrow (m-1)^2=-1\\\\\Rightarrow m-1=\pm√(-1)\\\\\Rightarrow m=1\pm i~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{where }i^2=-1]`

So, we get

`m_1=-1,~~m_2=1+i,~~m_3=1-i.`

Therefore, the general solution of the given equation is given by

`y(x)=Ae^(m_1x)+e^(1* x)(B\cos 1x+C\sin 1x)}\\\\\Rightarrow y(x)=Ae^(-x)+e^x(B\cos x+C\sin x).`

Thus, the required general solution is

`y(x)=Ae^(-x)+e^x(B\cos x+C\sin x).`