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 May 13th, 2013, 05:27 AM #1 Member   Joined: Apr 2013 Posts: 42 Thanks: 1 differential equation Given is the differential equation $\frac{1}{2b^{2}}{y}''(x)+b{y}'(x)+b^{4 }y(x)=0$ where b is constant and real. How do I show that for any $b> 0$ the solution to the equation is equal to $e^{-nx}\cdot (A\cos nx +B\cos nx)$, where $n=b^{3}$ and A and B are constant and real? And I also wonder how to determine for b=2, y(0)=2 and y'(0)=8 the solution to the differential equation?
 May 13th, 2013, 05:36 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: differential equation I would multiply the ODE by $2b^2$, then write the characteristic equation and use the quadratic formula to determine the roots. What do you find?
 May 13th, 2013, 06:44 AM #3 Member   Joined: Apr 2013 Posts: 42 Thanks: 1 Re: differential equation I find $r=-b^{3}\pm\sqrt{-b^{6}}=-b^{3}\pm b^{6}i$. So $y=e^{-b^{3}}[(C_{1}+C_{2})\cos b^{6}x+i(C_{1}-C_{2})\sin b^{6}x]=e^{-b^{3}}(A\cos b^{6}x+B\sin b^{6}x)$? Almost there, only I have a $b^{6}$ instead of $b^{3}$ in the sinus and cosinus...
 May 13th, 2013, 07:26 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: differential equation That is a second order, linear homogeneous differential equation. The theory says that the set of all solutions is a vector space of dimension 2 which means that every solution can be written as a linear combination of two functions that satisfy the equation. Clearly $e^{-nx}(Acos(nx)+ Bsin(nx))$ is a linear combination of $e^{-nx}cos(nx)$ and $e^{nx}sin(nx)$. So it is sufficient to show that $e^{-nx}cos(nx)$ and $e^{-nx}sin(nx)$ satisfy the differential equation and are independent.
 May 13th, 2013, 08:47 AM #5 Member   Joined: Apr 2013 Posts: 42 Thanks: 1 Re: differential equation I forgot to take the square root of $b^{6}$. When I do that, the answer is correct. And the second question is just substituting b by 2 and equate y and y' to the given values? It gives me A=2 and B=5/4

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