Question

Question #431e3

Answer 1

`e^(2x)y+3/2=(3e^(2x))/2`

Explanation:

`dy/(dx)+2y=3`

Can be expressed as:
`dy/dx+yf(x)=g(x)`

Where `f(x)=2` and `g(x)=3`.

Finding the Integrating Factor:
`int f(x) dx=int 2 dx=2x`

Raise `e` by this: `e^(2x)`
Multiply the original expression by this factor:
`dy/(dx)(e^(2x))+2e^(2x)y=3e^(2x)`

Observing the left hand side it is just the Product Rule applied.
Which means it can be rewritten as this:
`(e^(2x)y)'=3e^(2x)`

Integrate both sides
`int(e^(2x)y)'=int3e^(2x)`
`e^(2x)y+C=(3e^(2x))/2`

Find `C` given the condition `f(0)=0`.
`e^(2(0))(0)+C=3/2(e^(2(0)))`
`C=3/2`

Putting it all together:
`e^(2x)y+3/2=(3e^(2x))/2`