Question

1) Solve the following ODE with IVP 2y" + 6y' - 8y = 0 y(0) = 4 y'(0) = -1

Answer 1

Given differential equation ,  

2y" + 6y - Sy=0

Let   be a solution of the above differential equation .

y = e

Now ,  

y = e

y = memp, y" = memo

Substituting these values in the given differential equation we get ,

  

2m2 ету + бmenx — 8em = 0

→ (2m² + 6m - 8)eme = 0

→ 2m² + 6m - 8=0
, Since  $\small e^{mx}\neq 0$ .

$\small \Rightarrow 2(m^{2}+3m-4)=0$

$\small \Rightarrow m^{2}+3m-4=0$

$\small \Rightarrow (m+4)(m-1)=0$

$\small \Rightarrow m=-4,1$

So the solution of the given differential equation is ,

$\small y(t)=c_{1}e^{-4t}+c_{2}e^{t}$ where $\small c_{1},c_{2}$ are arbitrary constant and assuming  $\small t$ as independent constant , if the independent variable is  $\small x$ you can replace $\small t$ by  $\small x$ .

Now we will use initial conditions $\small y(0)=4$ and  $\small y'(0)=-1$ to find the value of the constants  $\small c_{1},c_{2}$ .

$\small y(0)=4$

$\small \Rightarrow c_{1}+c_{2}=4$$\small ......(i)$

$\small y'(0)=-1$

$\small \Rightarrow -4c_{1}+c_{2}=-1$   $\small ......(ii)$

$\small (i)-(ii)$ gives ,

$\small 5c_{1}=5$

$\small \Rightarrow c_{1}=1$

Substituting value of  $\small c_{1}$ in  $\small (i)$ we get ,

$\small 1+c_{2}=4$

$\small \Rightarrow c_{2}=3$

Substituting value of  $\small c_{1},c_{2}$ the reqired solution of the given differential equation is ,

Answer :  $\small y(t)=e^{-4t}+3e^{t}$ .

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