Question

Differential equation

Answer 1

Use the substitution
\[\frac{ dy }{ dx}=p\]
Where p is a function of x

To solve the differential equation

\[\frac{ d^2y }{ dx^2 }+\frac{ dy }{ dx}=1\]
Given that \[y=\frac{ dy }{ dx}=2 \] when x= -1

Answer 2

Its been a long time
I think you can begin this by writing d^2 y / dx^2 = dp/dx so

dp/dx + p = 1

Answer 3

hmm.. I'd have to revise this...

Answer 4

the general solution of the last equation is
p(x) = Ce^-x + 1 where x is some constant

Answer 5

so if we plug in p = dy/dx = 2 and x= -1 we can find the constant C.

Answer 6

to be honest I'm not 100% sure of this as its been a long time.

Answer 7

\(\dfrac{dy}{dx}=p\\\dfrac{d^2y}{dx^2}=\dfrac{dp}{dx}\)
So that the ODE becomes \(p'+p=1~~or~~p'+p-1=0\)
Now, you solve the first order of an ODE, it is easy, right?
The condition is \(p(-1) =2\)

Answer 8

The original 2nd order problem can be solved using "the usual methods"
So I am guessing that the only tool we have available is "separation of variables" ?
In other words, after making the substitution, you have
\[ \frac{dp}{dx} + p = 1 \\ \frac{dp}{dx} = 1-p \\ \frac{dp}{1-p}= dx \]
integration gives welsh's result (up above)
and you use the initial conditions p=2 when x=-1 to solve for the constant of integration
ultimately you get
\[ p= 1+e^{-x-1} \]
or
\[ \frac{dy}{dx} = 1+e^{-x-1} \\
dy= 1+e^{-x-1} \ dx\]
now integrate again. and solve for the constant of integration , using y=2 when x= -1

Answer 9

Thanks guys! Appreciate it :)