Question

find the particular solution that satisfies the initial conditions
f''(x)= sinx + e^2x
f(0)= 1/4, f'(0) =1/2

Answer 1

How much have you done with this problem so far?

Answer 2

Or haven't found the right place to begin? :)

Answer 3

@pnp97 what exactly u want?

Answer 4

i was sick for almost the whole chapter so i have no idea how to begin @rvc @AccessDenied

Answer 5

derivative of sinx?

Answer 6

Hm.. alright.
We are given the second derivative of a function. Our goal is to find the original function given that alone and some initial conditions (basically, just points that satisfy the equation).

To get back from the derivative, we need integration.

Answer 7

so take the integral of sinx+ e^2x??

Answer 8

Yep.
\( \dfrac{d^2}{dx^2} f(x) = \sin x + e^{2x} \)
\( \dfrac{d}{dx} \left( \dfrac{df}{dx} \right) = \sin x + e^{2x} \)
We can integrate both sides with respect to x to cancel the first derivative.

Answer 9

integral of sinx is -cosx and would e^2x stay the same??

Answer 10

As an example I just made up on the spot:

conditions:
f(1)=5
f'(0)=-2

differential equation:
f''(x)=2x

So integrate both sides you get:

f'(x)=x^2+C

since f'(0)=-2 then
-2=C

f'(x)=x^2-2

integrate again!

\[f(x)=\frac{ 1 }{ 3 }x^3-2x+C\]
Since f(1)=5 \[5=\frac{ 1 }{ 3 }(1)^3-2(1)+C\]
solve for the constant again and then plug it in and you get your final, particular equation of f((x).

if you didn't have the conditions you wouldn't have solved for the constants so you'd have C1 and C2 which would be the General solution rather than the Particular one.

Answer 11

Almost. If you took the derivative, d/dx e^(2x) = 2 e^(2x). You'll need to divide by 2 when integrating, because then by chain rule it works out when taking the derivative again as a check .

Answer 12

so you have to integrate twice? @kainui

Answer 13

ah so e^2x /2

Answer 14

Yes. Two integrations are required in total, this process is the first of the two, which we will add a constant C.

And correct! :)

Answer 15

So you have this equation now
f'(x) = -cos x + e^(2x) / 2
Notice your initial conditions now include one for f'(x). f'(0) =1/2

Answer 16

f'(x) = -cos x + e^(2x)/2 + C
wow i forgot the constant, bad me. That constnt is important!

Answer 17

The goal of that initial condition now is to solve for the constant. We can plug in x=0 and have f'(0) on one side, and the computed right side will become a 1-"variable" equation with C.

Answer 18

so plug in 0 or 1/2??

Answer 19

f'(x) = -cos x + e^(2x) / 2 + C <-- f'(0) = 1/2
f'(0) = -cos (0) + e^(2)0)) / 2 + C
^ ^ we can compute this part
this is 1/2

Answer 20

so the e^0 would cancel out and what would become of -cos(0)???

Answer 21

-cos (0)
You know the unit circle? The x-value of the angle 0 is the full radius.

and e^(0) is equal to 1, not 0. Easy thing to overlook. :)

Answer 22

oh so -1 + 1/2 +c!

Answer 23

Yes! :)
We have the equality to deal with here then:
1/2 = -1 + 1/2 + C
Solve for C, and we're half way there.

Answer 24

1=c and now what happens??/

Answer 25

It goes back into the equation:
f'(x) = -cos x + e^(2x) / 2 + 1 <--- C = 1 now.
and, for the most part, we do the whole process again, starting by taking the integral of both sides.

Answer 26

But next time, we're going to use that initial condition of f(0) = 1/4, because we will then have the function f alone without any derivatives.

Answer 27

so take the integral again and then plug in 0 again?

Answer 28

Pretty much, yep.
If the initial condition were for some reason f(1), we'd plug in x=1 to find the constant, but this case it is made easy. to use x=0

Answer 29

so sinx+e^2x +0?

Answer 30

that would be the derivative, getting back to the second derivative.

Answer 31

oh right itd be -sinx +e^2x+1 /3 +x?

Answer 32

hmm.. I agree with -sin x.
when we integrate e^(2x) / 2 again, that constant is kept. 1/2 times the integral of e^(2x) dx. you already took that integral before, remember what you had then?

Answer 33

\( \displaystyle \int \dfrac{e^{2x}}{2} \ dx = \frac{1}{2} \int e^{2x} \ dx \)
just to illustrate the coefficient

Answer 34

e^2x /2?

Answer 35

yep. that was the integral of e^(2x). and then we just multiply back the 1/2 in front:
1/2 * (e^(2x)/2)
makes sense?

Answer 36

can we write it as e^2x /4?

Answer 37

yes., that would be fine.
so: f(x) = -sin x + e^(2x) / 4 + .... just a bit more work
when we integrate a constant, it gains an x factor (a sort of reverse power rule, the derivative of x is 1.. so the integral of 1 is x)
\( \int 1 dx = x + C \)

Answer 38

twas what i said before! haha but without the c. so the integral is -sinx + e^2x /4 + x +c?

Answer 39

yep, that's right.
sorry, saw the + 1/3 + x and wasn't quite sure. :p

anyways, you have f(x) = -sin x + e^(2x) / 4 + x + C and now you can use f(0) = 1/4 to find the last constant.

Answer 40

no worries! is the answer c=0???

Answer 41

Yep. So your final answer is f(x) when you plug C = 0 back in.

Answer 42

its not done!!!!!!!!???? right so would i plug c=0 into the original sinx+e^2x?

Answer 43

into f(x) = -sin x + e^(2x)/4 + x + C <--- that C = 0
THEN you're done! :p

Answer 44

Basically, this f(x) is one particular solution to the original equation that satisfies those two initial conditions. If you took the derivative twice, you 'd get the original equation again.

Answer 45

AH TOOK 5EVER BUT THANKS SO MUCH:-) could you help me out on more equations?

Answer 46

Uhhhh weird, my old post is there ???
Sorry, it is getting late ! (1:26 AM)

I am sure there are others eager to help, though! :D

Answer 47

These problems do tend to take a while, but if you understand the process and get good practice with it, it can go a lot faster. :D

Answer 48

its ok no worries! i have a test on this soon so thank u for reverything!!1