1. ## Initial value problem

Find the particular solution of the initial value problem.

(a) dy/dx=e^-3x+2x; y(0)=5/3

(b) given that y=sin(3x+4y) find dy/dx

I have a test on monday and a topis similar to this will be included. I would really appreciate it if somebody could show me the workings and help me understand the math, so i can try some examples by myself.

Thanks alott!!

2. Hi jameslynskey

For (a) :
$\displaystyle \frac{dy}{dx}=e^{-3x}+2x$

$\displaystyle dy=(e^{-3x}+2x)dx$

Integrate both sides. Information : y(0)=5/3 will be used to find the value of constant resulting from the integration.

For (b) :
You have to use implicit differentiation to find $\displaystyle \frac{dy}{dx}$. Do you know it ?

I'm a little bit confused. Is the question asking to find the particular solution for question (b) ?

3. could you please show me the working throught the integration method aswell please?

And for part (b), i have worked through the 'Straud' mathematics book and figured it out (:

Thanks

4. Hi jameslynskey

$\displaystyle dy=(e^{-3x}+2x)dx$

$\displaystyle \int dy=\int (e^{-3x}+2x)dx$

$\displaystyle y=\int e^{-3x} \; dx \; + \int 2x \; dx$

I'm sure it's easy for you to find $\displaystyle \int 2x \; dx$

As for $\displaystyle \int e^{-3x} \; dx$ , can you find $\displaystyle \int e^x\; dx$ ?

5. Thanks alot. I really appreachaite the help.

One last question, what do i now have to do with the y(0)=5/3?

Btw i am only a high school student i am just trying really hard so i can do well at A level.

*Press Thanks button*

6. Hi jameslynskey

Please show me your integration result first because it will be easier to point out how to use y(0)=5/3

P.S. I will press thanks button after you completely finish with this problem hahahaha

7. Iv only been doing this for 2 days so beware lol...

y=e^-3x+x^2

and i meant i pressed thanks for you!

8. Hi jameslynskey

Let :

$\displaystyle u = -3x$

$\displaystyle \frac{du}{dx}= \; -3$

$\displaystyle dx=\frac{du}{-3}$

So, $\displaystyle \int e^{-3x} \; dx$

$\displaystyle =\int e^u\; \frac{du}{-3}$

$\displaystyle =-\frac{1}{3}\; \int e^u\; du$

Since $\displaystyle \int e^x\; dx\; =\; e^x$ , then $\displaystyle \int e^u\; du\; =\; e^u$

Hence, $\displaystyle \int e^{-3x} \; dx\; =\; -\frac{1}{3}\; e^{-3x}$

The integration will be : $\displaystyle y=-\frac{1}{3}\; e^{-3x}+x^2+c$ , where c is constant of integration

Using information y(0) = 5/3, we can find the value of c. We substitute the value of x = 0 and y = 5/3 to find c. Can you find c ?

P.S. I'm really going to press thanks button for you because of your effort

9. y=-1/3e^-3x+x^2+c

c=-1/3e^-3x+x^2-y ??

c=-1/3e^-3(0)+(0)^2-5/3

c=-2??

10. Hi jameslynskey
Originally Posted by jameslynskey

c=-1/3e^-3x+x^2-y ??
This is wrong. This should be : $\displaystyle c = y+\frac{1}{3}e^{3x}-x^2$

11. c=y+1/3e^3x-x^2

c=5/3+1/3e^3(0)-(0)^2

c=2 ???