1. ## differential equations

need help with this

e^3y dy/dx = cos2x + x^3

intergral of e^3y dy = intergral of cos2x + x^3 dx

1/3e^3y = 1/2sin2x + 1/4x^4 + C

how do i get y by itself

2. Originally Posted by mark18
need help with this

e^3y dy/dx = cos2x + x^3

intergral of e^3y dy = intergral of cos2x + x^3 dx

1/3e^3y = 1/2sin2x + 1/4x^4 + C

how do i get y by itself
multiply both sides by three to get:

$\displaystyle e^{3y}=\tfrac{3}{2}\sin(2x)+\tfrac{3}{4}x^4+C$

Now, take the natural log of both sides to get:

$\displaystyle 3y=\ln\left(\tfrac{3}{2}\sin(2x)+\tfrac{3}{4}x^4+C \right)$

Divide both sides by three to see that $\displaystyle \color{red}\boxed{y=\tfrac{1}{3}\ln\left(\tfrac{3} {2}\sin(2x)+\tfrac{3}{4}x^4+C\right)}$

Does this make sense?

--Chris

3. i knew that u could do e^(lnx) = x
didnt know u could do ln(e^x) = x
cheers mate

4. i get
y = 1/3 ln(3/2sin2x + 3/4x^4 + 3C)

or can i take the 3 out from inside the brackets to get

y = ln(1/2sin2x + 1/4x^4 + C)

?????

5. Originally Posted by mark18
i get
y = 1/3 ln(3/2sin2x + 3/4x^4 + 3C)

or can i take the 3 out from inside the brackets to get

y = ln(1/2sin2x + 1/4x^4 + C)

?????

You can't do that. It's illegal.

You're answer would remain as $\displaystyle y = \tfrac{1}{3} \ln\left(\tfrac{3}{2}\sin(2x) + \tfrac{3}{4}x^4 + C\right)$.

OR, you can bring the 1/3 into the natural log [as an exponent!] and rewrite it as $\displaystyle y =\ln\left(\sqrt[3]{\tfrac{3}{2}\sin(2x) + \tfrac{3}{4}x^4 + C}\right)$

--Chris

6. it should be 3C inside the brackets shouldnt it

7. Originally Posted by mark18
it should be 3C inside the brackets shouldnt it
"Yes", but 3C is just another constant. So we can call it C, or K...as long as we know that its a constant.

--Chris