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
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