Separable Equations

**dreamx20** · August 11th 2010, 01:20 AM

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

**Chris L T521** · August 11th 2010, 01:44 AM

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 $e^{a+b}=e^ae^b$ , not $e^a+e^b$ ...this would then allow you to separate the variables.

**helopticor** · August 14th 2010, 03:47 PM

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.

**Prove It** · August 14th 2010, 09:13 PM

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

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

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

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

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

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

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

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

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

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

**Prove It** · August 14th 2010, 09:17 PM

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

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

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

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

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

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

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

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

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

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

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

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