Question

Differential Equation y'-y=2x-3 solve with as a first order equation

Answer 1

The final answer is 2x+y-1=ce^x

Answer 2

Any thoughts fellas?

Answer 3

@Godlovesme ?

Answer 4

@Justin_Lee ?

Answer 5

Wasn't what I got. How did you get that answer?

Answer 6

from the book

Answer 7

Sorry. You are right. I copied down the original equation wrong when I was solving it. I thought it was y'+y.

Answer 8

Did you actually solve it?!

Answer 9

@Justin_Lee

Answer 10

Yeah. Have you taken differential equations before?

Answer 11

not at all

Answer 12

What level of math are you?

Answer 13

tbh it was only our second lesson

Answer 14

Okay. Do you know the method of integrating factors?

Answer 15

Sorry. I haven't eaten in a while, so I was getting food.

Answer 16

sort of, and first for it, boun appetite

Answer 17

with all due resfect, it's more important

Answer 18

respect

Answer 19

Thanks. Okay. This can be solved using either the method of integrating factors or using a characteristic equation.

Answer 20

The method of integrating factor requires you to multiply the entire equation by a dummy expression, say u(x).

Answer 21

This gives you:
y'(x)*u(x) - y(x)*u(x) = 2x*u(x) - 3u(x).

Answer 22

This allows you to take advantage of the product rule for the left hand side of the equation. Remember, the product rule is: (fg)' = g'f + f'g.

Answer 23

That now kind of resembles the left hand side of the expression: y'(x)*u(x) - y(x)*u(x). This assumes that u(x) = f and y(x) = g.

Now, notice something interesting here. u'(x) = u(x), if we are following the pattern set by the product rule.

Answer 24

You can recognize that only one function can do this: e^x. Or, you can prove this to yourself in the following manner:

u'(x) = du(x)/dx = u(x)

du(x)/u(x) = dx

Integrating both sides gives:
ln(u(x)) = x (the constant is irrelevant)
Raising both sides by an exponential yields:
u(x) = e^x.

Answer 25

Now, you can simplify the left hand side from: y'(x)*u(x) - y(x)*u(x) to: (y(x)*u(x))'
by following the steps backwards from the product rule: g'f + f'g = (fg)'

Keep in mind that u(x) = e^x.

Answer 26

Does all this make sense so far?

Also, I made a mistake in the second to last post. Can you spot it?

Answer 27

no, can you tell me where is it? sorry it's too late here to think that hard lol

Answer 28

but @Justin_Lee I'll look at that in the morning thank you so much!

Answer 29

such a vivid explanation!

Answer 30

u(x) should be e^-x.

Answer 31

Sure. I will type up some concluding remarks and let you be on your way.

Answer 32

wow! Truly thanks a lot!!!!!!

Answer 33

It has been a long day

Answer 34

u(x) should be e^-x because of the y' - y.

Imagine that u(x) = e^ax. Then, if we multiply some y(x) by u(x) and take the derivative of that product, we would yield: y'(x)*u(x) + a*y(x)*u(x) [prove it to yourself].

Now, having simplified the expression by reversing the product rule, we have:
(y(x)*u(x))' = 2x*u(x) - 3*u(x).

Replace u(x) with e^-x gives:
(y(x)*e^-x)' = 2x*e^-x - 3*e^-x

Integrate both sides by x to get rid of the derivative on the left hand side (don't forget to tack on a constant to the right hand side). This gives:
y(x)*e^-x = -2x*e^-x + e^-x + c [solve it out to prove it to yourself that this is the case]

Divide by e^-x in both sides to get the answer that you found in the book.

Answer 35

Alternatively, you can use the DE's characteristic equation in conjunction with the method of undetermined coefficients (which is what people usually use to solve these types of problems). This method sometimes requires a leap of faith at first in order to believe your result, but is much faster than the other method I showed you. If you are working with multiple derivatives, expect to learn this in the near future.