Question

Solve the differential equation.
y ' = 9x - y

Answer 1

I would write it in standard form
y' - y = 9x

first find the homogenous solution :
y' - y = 0

do you know how to do that ?

Answer 2

No

Answer 3

you assume the solution has the form
y= e^(ax)
sub that into the equation. what do you get ?

Answer 4

\[y ' + e^{ax} = 0\]
?

Answer 5

yes, but if y= e^ax
you can find y'

Answer 6

what is d/dx of e^(ax) ?

Answer 7

\[ae^{ax}\]

Answer 8

yes, put that in for y'

Answer 9

\[ae^{ax}+e^{ax} = 0\] ?

Answer 10

yes
e^ax is not zero (except at plus or minus infinity depending on the sign of a)
so we can divide both sides by e^ax
then solve for a

Answer 11

\[a + 1 = 0\] ?

Answer 12

\[\frac{ ae^{ax} }{ e^{ax} }+\frac{ e^{ax} }{ e^{ax} } = 0\]

Answer 13

yes

a+1=0
is called the characteristic equation. people usually "cut to the chase"
and write it down immediately:
y' + y =0 becomes a+1 = 0
(another example: y'' + 2y' -3y would become a^2 +2a -3 = 0 )

anyway, the homogenous solution is
y = c e^(ax) with a=1:

y = c e^x

Answer 14

next we need to find the "particular solution"
i.e. the solution that solves
y' + y = 9x

Answer 15

one strategy is to *assume* the solution has the same form as the right-hand side, and form a linear combination of it and its derivatives

what that means is we assume the solution is Ax (A is an unknown coefficient)
and the derivative of x: 1, but scaled by another unknown coefficient

in other words, we try
y = Ax + B

put that into the equation
y' + y = 9x
what do you get ?

Answer 16

\[y' + Ax + B = 9x\]?

Answer 17

yes for y. do the same for y' (first find d/dx of y)

Answer 18

\[\frac{ dy }{ dx }(Ax+B) = A, y ' = A\]\[A + Ax+B = 9x\]
?

Answer 19

yes, now match up terms
Ax + (A+B) = 9 x + 0

Answer 20

A matches with 9
A+B with 0

Answer 21

Okay

Answer 22

we know A, what is B?

Answer 23

B = - 9? Since Ax/x = 9.

Answer 24

Ax + (A+B) = 9x + 0
the coeff in front the x's must match up. so A=9
the constant A+B = 0
with A=9 we have 9+B= 0
B= -9

Answer 25

that makes the particular solution
y= 9x - 9

the homogenous solution was from
a+1 = 0
a= -1
so y= c e^-x (I make it e^x up above, but that was wrong)

the total solution is the sum of the homogenous and particular solutions

Answer 26

\[ y = c e^{-x} + 9x - 9 \]
where c is an unknown constant. If we have initial conditions we can find its value.

Answer 27

\[y=ce^{-x}+9x-9\] would be the answer if no initial conditions are given?

Answer 28

yes. we can check it in the original equation
first find y'

Answer 29

\[y' = -ce^{-x}+9\]

Answer 30

the other side is
9x - y = 9x - c e^-x -9x + 9
or
-c^-x +9 (the 9x - 9x cancel)
which matches y' (the left side)

Answer 31

Thanks for the walk-through! :)