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 April 9th, 2010, 08:50 AM #1 Senior Member   Joined: Jan 2009 Posts: 344 Thanks: 3 find m and b where y=mx+b is a solution to dy/dx Consider the differential equation $\frac{dy}{dx}= \frac{1}{2}x+y-1$ Find the values of the constants m and b, for which $y= mx+b$ is a solution to the differential equation? Correct me if I'm wrong. Now understand that $\frac{dy}{dx}= m$ substitute $(mx+b)$ for $y$ in $\frac{dy}{dx}= \frac{1}{2}x+y-1$ so $m= \frac{1}{2}x + (mx+b)-1$ $m= (m+\frac{1}{2})x + (b-1)$ Since there is no x term at the left hand side of the equation, the coefficient of x at the right-hand side of the equation must be equal to Zero  thus $m= -\frac{1}{2}$ Making $-\frac{1}{2}=b-1$ thus $b=\frac{1}{2}$
 April 9th, 2010, 10:19 AM #2 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: find m and b where y=mx+b is a solution to dy/dx Sure. So what do you have so far?
 April 9th, 2010, 12:28 PM #3 Senior Member   Joined: Nov 2008 Posts: 265 Thanks: 0 Re: find m and b where y=mx+b is a solution to dy/dx you need some initial value to go ahead the way I see it. And please, as jason.spade implied, at least make an educated guess. You'll learn better that way if you get the idea but just couldn't get the answer, or whatever way.
 April 10th, 2010, 12:55 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,547 Thanks: 1477 Any solution of the equation can be written as $\small{\text{y}\,=\,(-1/2)\text{x}\,+\,1/2\,+\,(y_0\,-\,1/2)e^{\text{x}},}$ where $\small{\text{y_0}}$ is the value of y when x = 0.
April 12th, 2010, 05:50 PM   #5
Senior Member

Joined: Jan 2009

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Re:

Quote:
 Originally Posted by skipjack Any solution of the equation can be written as $\small{\text{y}\,=\,(-1/2)\text{x}\,+\,1/2\,+\,(y_0\,-\,1/2)e^{\text{x}},}$ where $\small{\text{y_0}}$ is the value of y when x = 0.
Thats not the answer as far as this problem is concerned.
Nice work though.

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# the value of constant m and c for which y= mx c is a solution of the differential equation

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