Question

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

Answer 1

Answer: The required solution is

`y(t)=-(7)/(3)e^(-t)+(7)/(3)e^{(1)/(5)t}.`

Step-by-step explanation: We are given to solve the following differential equation :

`5y^(\prime\prime)+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)`

Let us consider that

`y=e^(mt)` be an auxiliary solution of equation (i).

Then, we have

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

Substituting these values in equation (i), we get

`5m^2e^(mt)+3me^(mt)-2e^(mt)=0\\\\\Rightarrow (5m^2+3y-2)e^(mt)=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^(mt)\\eq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~(1)/(5).`

So, the general solution of the given equation is

`y(t)=Ae^(-t)+Be^{(1)/(5)t}.`

Differentiating with respect to t, we get

`y^\prime(t)=-Ae^(-t)+(B)/(5)e^{(1)/(5)t}.`

According to the given conditions, we have

`y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)`

and

`y^\prime(0)=2.8\\\\\Rightarrow -A+(B)/(5)=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-(14)/(6)\\\\\Rightarrow A=-(7)/(3).`

From equation (ii), we get

`B=(7)/(3).`

Thus, the required solution is

`y(t)=-(7)/(3)e^(-t)+(7)/(3)e^{(1)/(5)t}.`