Differential Equations - maximal interval

[Paper Wings]

Homework Statement

1. Write an interval formula for the solution
$$f'(x)=2f(x)+e^x$$
f(1)=0

Explicitly find the maximal interval I about 0 on which we can solve the following differential equations
2. f'(x) = xf(x)
f(0)=1

3. $$f'(x)=[f(x)]^2$$
f(0)=-1

Homework Equations

For
a.$$f'(x)=bf(x)+h(x)$$
b. $$f(x_{0})=y_{0}$$
then
$$e^{b(x-x_{0})}+ \int{e^{b(x-t)}h(t)dt}$$

The Attempt at a Solution

1. By applying the formula, it yields
$$f(x)=e^{2(x-1)}0+ \int_0^x{e^{2(x-t)}dt}=\left[ -\frac{1}{3}e^{3x-2t}\right]\right|^{x}_{1}= \left[ -\frac{1}{3}e^x- \left(- \frac{1}{3}e^{3x-2} \right) \right]$$

2. I tried dividing f(x) from both sides of the f'(x) equation then I know that $$\frac{f'(x)}{f(x)}= \frac{d}{dx} \left[ lnf(x) \right]$$, but I do not know how to procede from here.

3. I've tried using the same method with part 2 except that I divide one f(x) from both sides which yields $$\frac{f'(x)}{f(x)}=f(x)$$. Again I'm stuck. Any hints, tips, or help is greatly appreciated. Thank you.

 

[foxjwill]

alright, let's see.

For 1), you accidentally changed your lower bound from 0 to 1. Also, for some reason, you multiplied $$e^{2(x-1)}$$ by 0, which you should not have done.

Both 2) and 3) are simple separation of variables problem. However, I'm not entirely sure what the question means by "maximal interval". The maximum with respect to x? If so, than you don't even need to solve the differential equation. Instead just solve for x and determine whether or not its increasing or decreasing at those points.