# Eliminating the arbitrary constant

• Feb 11th 2011, 08:46 AM
sparky
Eliminating the arbitrary constant
Hi,

I am having trouble understanding the steps needed to eliminate the arbitrary constants from the equation y = Ax + 3. The answer in the book is y = x (dy/dx) + 3. Here are my steps along with my reasoning for each step (please correct my reasoning so I can finally understand how to do this):

y = Ax + 3

dy/dx = A (I am still left with A, so I need to find another way to get rid of A)

y = Ax + 3 (starting over)

(y - 3)/x = A (isolate A)

(y - 3)/x = dy/dx (replace A with dy/dx - is this correct? if so, why?)

y - 3 = (dy/dx)x

y = (dy/dx)x + 3 (isolate y. Now we are left with an equation that does not have A in it )
• Feb 11th 2011, 08:49 AM
Ackbeet
Solve for A, differentiate both sides using implicit differentiation. HT: Danny.
• Feb 11th 2011, 09:21 AM
topsquark
Quote:

Originally Posted by sparky
Hi,

I am having trouble understanding the steps needed to eliminate the arbitrary constants from the equation y = Ax + 3. The answer in the book is y = x (dy/dx) + 3. Here are my steps along with my reasoning for each step (please correct my reasoning so I can finally understand how to do this):

y = Ax + 3

dy/dx = A (I am still left with A, so I need to find another way to get rid of A)

y = Ax + 3 (starting over)

(y - 3)/x = A (isolate A)

(y - 3)/x = dy/dx (replace A with dy/dx - is this correct? if so, why?)

y - 3 = (dy/dx)x

y = (dy/dx)x + 3 (isolate y. Now we are left with an equation that does not have A in it )

Am I missing some detail here?
y = Ax + 3

dy/dx = A

Put this value of A into the original equation:
y = (dy/dx)x + 3

which is the desired solution.

-Dan
• Feb 11th 2011, 10:21 AM
Ackbeet
Quote:

Originally Posted by topsquark
Am I missing some detail here?
y = Ax + 3

dy/dx = A

Put this value of A into the original equation:
y = (dy/dx)x + 3

which is the desired solution.

-Dan

Certainly nothing wrong with this method, and it is the most direct. However, it is not very general. Danny's method will work in pretty much any case that has a single arbitrary constant for which you can solve (which is true for many first-order ODE's).