solve the differential equation

**differentiate** · January 15th 2010, 08:31 PM

Solve the differential equation dy/dx= 2y for y given that when x = 1, y = 1

Thanks

**Calculus26** · January 15th 2010, 08:54 PM

1. separate the variables

dy/y = 2dx

2. integrate

ln|y| = 2x+ c

3. Apply ICs x= y = 1

0 = 2 + c

c=-2

ln|y| = 2x - 2

4. exponentiate

y = e^(2x-2) = e^(-2) e^(2x)

**Prove It** · January 15th 2010, 08:57 PM

Originally Posted by differentiate

Solve the differential equation dy/dx= 2y for y given that when x = 1, y = 1

Thanks

$\frac{dy}{dx} = 2y$

Method 1:

$\frac{dx}{dy} = \frac{1}{2y}$

$x = \int{\frac{1}{2y}\,dy}$

$x = \frac{1}{2}\ln{|y|} + c$

$x - c = \frac{1}{2}\ln{|y|}$

$2x - C = \ln{|y|}$ where $C = 2c$

$|y| = e^{2x - C}$

$|y| = e^{-C} e^{2x}$

$y = \pm e^{-C} e^{2x}$

$y = A e^{2x}$ where $A = \pm e^{-C}$ .

Now letting $x = 1, y = 1$ we find

$1 = Ae^{2(1)}$

$1 = Ae^2$

$A = e^{-2}$ .

So $y = e^{-2} e^{2x}$

$y = e^{2x - 2}$ .

Other possible methods are separation of variables or integrating factor.

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