General solution of the differential

[hoplessatmaths]

Hi
I'm (very) stuck! I have to use an answer from a previous question (f'(x)=2/3cos(x) e^x(2/3sin(x)) to find the general solution of the differential equation:

dy/dx = 2cos x e^x (2/3sin x)y^2/3 (y>0)

Answer given in implicit form.

So far I've got

dy/y^2/3=2 cos x e^x(2/3sin x) dx

Can you please give me a push in the right direction?

 

[BigGlenntheHeavy]

$\displaystyle What \ is \ \frac{dy}{dx}?$

$\displaystyle You \ have \ \frac{dy}{dx} \ = \ (2)(cos(x))(e^{x})\bigg(\frac{2}{3sin(x)}\bigg)\bigg(\frac{y^{2}}{3}\bigg)$

$\displaystyle I \ suspect \ that \ isn't \ right.$

$\displaystyle If \ you \ want \ help \ from \ us, \ get \ it \ together.$

 

[galactus]

If you have $\displaystyle \frac{dy}{dx}=2cos(x)e^{x}\cdot \frac{2}{3}sin(x)\cdot \frac{y^{2}}{3}$

Then separate variables and integrate.

$\displaystyle \frac{3}{y^{2}}dy=\frac{4}{3}cos(x)e^{x}sin(x)dx$

$\displaystyle 3\int \frac{dy}{y^{2}}=\frac{2}{3}\int e^{x}sin(2x)dx$

Now integrate both sides remembering the integration constant.

 

[BigGlenntheHeavy]

Dear hopelessatmaths;

$\displaystyle (2/3sin(x)) \ = \ \frac{2sin(x)}{3} \ (is \ mathematically \ correct), \ however \ it \ behooves$

$\displaystyle one \ to \ use \ grouping \ symbols \ if \ only \ for \ emphasis, \ so \ we \ have \ a \ better \ grasp \ of \ what \ you$

$\displaystyle \ are \ saying, \ as \ [(2/3)sin(x)] \ leaves \ no \ doubt. \ Sorry \ if \ I \ was \ a \ little \ testy, \ I \ apologize.$

 

[hoplessatmaths]

Hi don't worry about the testy bit! I'm having a really bad day on integration, and everything else to do with calculus! Could you please give a little help on the integration process?

Many thanks

 

[Deleted member 4993]

Hi don't worry about the testy bit! I'm having a really bad day on integration, and everything else to do with calculus! Could you please give a little help on the integration process?

Many thanks

I'll do a similar but different integration for you (these are standard integration - should be in your textbook).

$\displaystyle \int e^x \cos(x) dx \, = \, e^x\cdot \sin(x) \, - \, \int e^x \sin(x) dx$

$\displaystyle \int e^x \cos(x) dx \, = \, e^x\cdot \sin(x) \, - \, [-e^x\cdot \cos(x) \, + \, \int e^x \cos(x) dx]$

$\displaystyle \int e^x \cos(x) dx \, = \, \frac{1}{2}\cdot [e^x\cdot \sin(x) \, + \, e^x\cdot \cos(x) ]$

Follow the same procedure...

 

[BigGlenntheHeavy]

Using galactus"s solution, we get;

$\displaystyle 3\int \frac{dy}{y^{2}} \ = \ \frac{2}{3}\int e^{x}sin(2x)dx$

$\displaystyle Hence, \ \frac{-3}{y} \ = \ \frac{2e^{x}[sin(2x)-2cos(2x)]}{15}+C, \ I \ by \ P.$

$\displaystyle \frac{-y}{3} \ = \ \frac{15}{2e^{x}[sin(2x)-2cos(2x)]}+C$

$\displaystyle y \ = \ -\frac{45}{2e^{x}[sin(2x)-2cos(2x)]}+C$