Question

solve this differential equation

Answer 1

first order linear differential equation :

\(y' + p(x) y = q(x) \)

Answer 2

start by cmputing the integrating factor : \(\large e^{\int p ~dx}\)

Answer 3

IF: e^-5x?

Answer 4

yes, multiply IF both sides

Answer 5

before that, arrange the equation in standard form

Answer 6

what do u mean by standard form?

Answer 7

below :

\(\large \mathbb{y'-5y = x}\)

Answer 8

now multiply IF both sides

Answer 9

hmm.. ok , (y'-5y)e^-5x = xe^-5x ?

Answer 10

yes, distribute left side

Answer 11

idea of multiplying IF both sides is to get the left side into form :

p'q + pq'

Answer 12

\(\large \mathbb{y'-5y = x}\)

multiply IF both sides

\(\large \mathbb{y'e^{-5x}-5ye^{-5x} = xe^{-5x}}\)

Answer 13

left side is a derivative of product of two functions, right ?

Answer 14

\(\large \mathbb{y'-5y = x}\)

multiply IF both sides

\(\large \mathbb{y'e^{-5x}-5ye^{-5x} = xe^{-5x}}\)


\(\large \mathbb{(ye^{-5x})' = xe^{-5x}}\)

Answer 15

yeah okay how do i proceed?

Answer 16

next integrate both sides

Answer 17

left side gives u back the product funciton,
right side u can use parts..

Answer 18

y' = x + 5y
y' - 5y = x

y=yc+yp

Where yc is the general solution of y' - 5y = 0 and yp is a particular solution of y' - 5y = x

Characteristic equation: m-5=0

Then, m=5 and yc=A*e^(5*x) ; A is a real number

To find yp we have to propose a posible solution (We always propose trying with a polynomial of the same degree as the one on the right of the equation)

So we propose yp=B*x+C , and we replace it in the differential equation.

(B*x+C)' - 5(B*x+C) = x

B-5Bx-5C=x
x(-5B)+(B-5C)=x

For those two polynomials to be the same, their coefficients have to be equal. So:

-5B=1 and B-5C=0

B=-1/5 and C=-1/25

Then, yp=B*x+C=-x/5-1/25

And finally, y=yc+yp=A*e^(5*x)-x/5-1/25

y=A*e^(5*x)-x/5-1/25

Answer 19

where did 5 go?

Answer 20

differentiate it and see wat u get

Answer 21

\(\large \mathbb{(ye^{-5x})' } = ? \)

Answer 22

aww.. product rule?

Answer 23

yess
multiplying by IF fixes the left hand side to product rule...
so that taking integral becomes trivial

Answer 24

okay if i differentiate right -5y(e^-5x)?

Answer 25

how did u get that ?

\(\large \mathbb{(ye^{-5x})' } = y'e^{-5x} + y(e^{-5x})'\)

Answer 26

\(\large \mathbb{y'-5y = x}\)

multiply IF both sides

\(\large \mathbb{y'e^{-5x}-5ye^{-5x} = xe^{-5x}}\)

grouping left hand side by product rule :

\(\large \mathbb{(ye^{-5x})' = xe^{-5x}}\)

Answer 27

ahh.. im kinda confused by the ' anyway, why do we make it to form p'q+pq'?

Answer 28

lets carry out the integration and u will see why maybe :)

Answer 29

hmm.. integration by parts? both sides?

Answer 30

\(\large \mathbb{y'-5y = x}\)

multiply IF both sides

\(\large \mathbb{y'e^{-5x}-5ye^{-5x} = xe^{-5x}}\)

grouping left hand side by product rule :

\(\large \mathbb{(ye^{-5x})' = xe^{-5x}}\)

Integrate both sides

\(\large \mathbb{\int (ye^{-5x})' dx = \int xe^{-5x}} dx\)

Answer 31

u need to recall the Fundamental Theorem of Calculus 2 :

\(\int f(x)' dx = f(x)\)

Answer 32

left hand side, u are taking integral-of-a-derivative of a function
so it gives u back the funciton

Answer 33

ohh so that means it ye^-5x in the end ?

Answer 34

yup ! and thats true upto a constant...
but dont wry about the details for now... go ahead taking the integral of right side also.. .

Answer 35

\(\large \mathbb{y'-5y = x}\)

multiply IF both sides

\(\large \mathbb{y'e^{-5x}-5ye^{-5x} = xe^{-5x}}\)

grouping left hand side by product rule :

\(\large \mathbb{(ye^{-5x})' = xe^{-5x}}\)

Integrate both sides

\(\large \mathbb{\int (ye^{-5x})' dx = \int xe^{-5x}} dx\)

\(\large \mathbb{ ye^{-5x} = \int xe^{-5x}} dx\)

Answer 36

ok.. RHS: -(1/5)xe^-5x - (1/25)e^-5x +c

Answer 37

lastly, isolate y, by dividing e^-5x both sides

Answer 38

y = -(1/5)x- (1/25) +Ce^5x

Answer 39

hmm.. i still dont get the p'q +pq' part, and how y'-5y got to ye^-5x in the first place..

Answer 40

(ye^-5x)'

Answer 41

good question :)

lets see what we get, by taking the derivative of that

Answer 42

(ye^-5x)' = ?

take the derivative and see wat u get

Answer 43

take the derivative with respect to x

Answer 44

y'e^-5x +y(e^-5x)' as you said just now?

Answer 45

take the derivative fully

Answer 46

y'e^-5x +y(e^-5x)'
--------

Answer 47

take the derivative of that also

Answer 48

-5e^-5x?

Answer 49

yes, so u have :
(ye^-5x)' = y'e^-5x - 5ye^-5x

Answer 50

now look at wat you're asking now...

Answer 51

when u multiply by IF, left hand turns to derivative of product magically... go thru the solution again... u wil understand i hope :)

Answer 52

\(\large \mathbb{y'e^{-5x}-5ye^{-5x} = xe^{-5x}}\)

grouping left hand side by product rule :

\(\large \mathbb{(ye^{-5x})' = xe^{-5x}}\)

Answer 53

those two steps are equivalent.

we have just used the product rule in reverse and grouped...

Answer 54

ohh.. reverse product rule.. i see now, thank you very muchh! XD

Answer 55

ahh nice name 'reverse product rule' lol never heard of it before ;)