Question

Solve the differential equation y' = 0.3y given the condition that y(0) = 9.

Answer 1

So uhh looks like this one is separable if you just wanna do it that way,\[\Large\rm \frac{dy}{dx}=0.3y\qquad\to\qquad \frac{dy}{y}=0.3dx\]Understand how to integrate each side?

Answer 2

Okay so what am I supposed to integrate?

Answer 3

There is a differential on each side, so integrate each side with respect to it's given variable.
\[\Large\rm \int\limits \frac{dy}{y}=\int\limits 0.3dx\]

Answer 4

I got 3x/10 +C

Answer 5

I don't think it is right though because that is not one of the answers I have.

Answer 6

I don't understand what you're saying...
You got that from integrating the right side?
Or the left..?
Or what..?

Answer 7

the right side

Answer 8

Oh .3x+C, I see.

And what about integrating the left side?
remember your integral for 1/x?

Answer 9

the integral for 1/x is ln(x)

Answer 10

right?

Answer 11

yes, good.
So integrating each side you should get:\[\Large\rm \ln y+c_1=0.3x+c_2\]The constants are arbitrary so let's combine them,
subtract c1 from each side, absorb it into the other c,\[\Large\rm \ln y=0.3x+C\]

Answer 12

Before solving for y, it might be a good idea to use your initial data to find your C value.

Answer 13

\[\Large\rm y(0)=9\quad means\quad y=9~when~x=0\]
Plug it in and do some stuff.
\[\Large\rm \ln 9=0.3(0)+C\]

Answer 14

I am finding C right now. Right?

Answer 15

k got 2.197

Answer 16

Uhhh yes, but let's leave it as ln9.

Answer 17

So your goal is to get this equation,
\[\Large\rm \ln y=0.3x+\ln9\]in terms of y=stuff.
See how the log is causing a problem?

Answer 18

yes how do we fix tht problem?

Answer 19

Exponentiate each side,
Rewrite each side as an exponent with a base of e,\[\Large\rm e^{\ln y}=e^{0.3x+\ln9}\]

Answer 20

Exponential and log are inverse functions of one another,
they simply "undo" one another,\[\Large\rm y=e^{0.3x+\ln9}\]But it can be simplified a little further,

Answer 21

Applying rules of exponents to the right side,\[\Large\rm y=e^{\ln9}\cdot e^{0.3x}\]Then you can apply the exponent/log tricky again.

Answer 22

=9e^0.3x

Answer 23

yes

Answer 24

Awesome! Thanks!