solve the differential equation

• Jan 15th 2010, 09:31 PM
differentiate
solve the differential equation
Solve the differential equation dy/dx= 2y for y given that when x = 1, y = 1

Thanks
• Jan 15th 2010, 09:54 PM
Calculus26
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)
• Jan 15th 2010, 09:57 PM
Prove It
Quote:

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.