# Separable Equations

• Aug 11th 2010, 01:20 AM
dreamx20
Separable Equations
The questions are read solve the differential equation

a) sec(x)(dy/dx)=e^(y+sinx)
b)dy/dx= e^(2x-y)/e^(x+y)

For a all i could do was separate the right side to e^y +e^sinx while for b) I simplified to e^(x-2y). I am trying to read the text which is not much help since the sample questions are a lot more easier. Please help
• Aug 11th 2010, 01:44 AM
Chris L T521
Quote:

Originally Posted by dreamx20
The questions are read solve the differential equation

a) sec(x)(dy/dx)=e^(y+sinx)
b)dy/dx= e^(2x-y)/e^(x+y)

For a all i could do was separate the right side to e^y +e^sinx while for b) I simplified to e^(x-2y). I am trying to read the text which is not much help since the sample questions are a lot more easier. Please help

Recall that $\displaystyle e^{a+b}=e^ae^b$, not $\displaystyle e^a+e^b$...this would then allow you to separate the variables.
• Aug 14th 2010, 03:47 PM
helopticor
If you're in some sort of calculus class, it might be a good idea to review your algebra with a site like Purplemath. You should probably know all of the "elementary algebra" topics and most of the "intermediate algebra" topics. You'll find exponentiation rules like the one Chris L T521 mentioned there.
• Aug 14th 2010, 09:13 PM
Prove It
Quote:

Originally Posted by dreamx20
The questions are read solve the differential equation

a) sec(x)(dy/dx)=e^(y+sinx)
b)dy/dx= e^(2x-y)/e^(x+y)

For a all i could do was separate the right side to e^y +e^sinx while for b) I simplified to e^(x-2y). I am trying to read the text which is not much help since the sample questions are a lot more easier. Please help

1) $\displaystyle \sec{x}\,\frac{dy}{dx} = e^{y + \sin{x}}$

$\displaystyle \sec{x}\,\frac{dy}{dx} = e^y\,e^{\sin{x}}$

$\displaystyle e^{-y}\,\frac{dy}{dx} = \cos{x}\,e^{\sin{x}}$

$\displaystyle \int{e^{-y}\,\frac{dy}{dx}\,dx} = \int{\cos{x}\,e^{\sin{x}}\,dx}$

$\displaystyle \int{e^{-y}\,dy} = \int{e^u\,du}$ if we make the substitution $\displaystyle u = \sin{x}$

$\displaystyle -e^{-y} + C_1 = e^u + C_2$

$\displaystyle -e^{-y} = e^{\sin{x}} + C$ where $\displaystyle C = C_2 - C_1$

$\displaystyle e^{-y} = -e^{\sin{x}} - C$

$\displaystyle -y = \ln{(-e^{\sin{x}} - C)}$

$\displaystyle y = -\ln{(-e^{\sin{x}} - C)}$.
• Aug 14th 2010, 09:17 PM
Prove It
Quote:

Originally Posted by dreamx20
The questions are read solve the differential equation

a) sec(x)(dy/dx)=e^(y+sinx)
b)dy/dx= e^(2x-y)/e^(x+y)

For a all i could do was separate the right side to e^y +e^sinx while for b) I simplified to e^(x-2y). I am trying to read the text which is not much help since the sample questions are a lot more easier. Please help

2) $\displaystyle \frac{dy}{dx} = \frac{e^{2x-y}}{e^{x+y}}$

$\displaystyle \frac{dy}{dx} = \frac{e^{2x}\,e^{-y}}{e^x\,e^y}$

$\displaystyle \frac{dy}{dx} = e^x\,e^{-2y}$

$\displaystyle e^{2y}\,\frac{dy}{dx} = e^x$

$\displaystyle \int{e^{2y}\,\frac{dy}{dx}\,dx} = \int{e^x\,dx}$

$\displaystyle \int{e^{2y}\,dy} = \int{e^x\,dx}$

$\displaystyle \frac{1}{2}e^{2y} + C_1 = e^x + C_2$

$\displaystyle \frac{1}{2}e^{2y} = e^x + C$ where $\displaystyle C = C_2 - C_1$

$\displaystyle e^{2y} = 2e^{x} + 2C$

$\displaystyle 2y = \ln{(2e^x + 2C)}$

$\displaystyle y = \frac{1}{2}\ln{(2e^x + 2C)}$.