how do i solve for y(x) if dy/dx=x+y and y(0)=1

Question

I know the answer is y(x)=2e^x-x-1

I posted this question before but i can't find my question so i would appreciate if you helped me again.

Answer 1

As BobPursley and I explained in our previous answer, what you have is a first order linear differential equation.

The general form of this type of differential equation is
dy/dx = P(x)*y + Q(x)
In your case,
P(x) = 1, and
Q(x) = x

The procedure for obtaining a general solution is explained here:
http://www.sosmath.com/diffeq/first/lineareq/lineareq.html

You need to calculate the integral:
r(x) = e^(integralP(x)dx)
which in your case is just r(x) = e^x

The solution is then
r(x)*y(x) = integral[r(x)*Q(x)] dx + C
which can be rewritten
e^x*y(x) = integral[x*e^x] dx + C

A few more steps will lead you to the answer. The constant C needs to be selelected to fit your initial value y(0)

There should be a section of your text that provides examples of using this technique.

You really need to attempt this yourself. Show your work if you need additional help.