# Find the solution

• Aug 19th 2009, 12:42 PM
SmallMan
Find the solution
a) Find the solutnion of the following differential equation:

(d^2y)/(dx^2)+2dy/dx-8y=0

b) Find the solutnion of the following differential equation:

(d^2y)/(dx^2)+2dy/dx-8y=14e^3x

With initial conditions: at x=0, y(0)=3 and (dy/dx)(0)=-4

My last problem, i tried to solve but every time i was getting other solutions. Can somebody tell me (especially) explain how to solve this ?
• Aug 19th 2009, 01:49 PM
Mauritzvdworm
$\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}-8y=0$

can be rewritten in the following way

$r^{2}+2r-8=0$

this we can factorise as follows

$(r+4)(r-2)=0$

so the equation has roots at -4 and 2 hence the answer to the homogenous DE is

$y_{homogenous}=Ae^{-4x}+Be^{2x}$

where A and B are both constants

the second problem is almost the same, we have already determined the solution to the homogenous equation which had the roots mentioned above. Since the term on the right hand side $14e^{3x}$ does not correspondto any of the roots we can write the solution according to the superposition principle as follows

$y_{general}=Ae^{-4x}+Be^{2x}+Ce^{3x}$

we can now also find C since we know the homogenous solution to yield zero we only need to do the same with the non homogenous solution. So plug $CE^{3x}$ into the differential equation and solve for C. You should find C=2, then you can write the general solution as follows

$y_{general}=Ae^{-4x}+Be^{2x}+2e^{3x}$

using the initial conditions you can then solve for A and B by just plugging them in.
• Aug 19th 2009, 01:51 PM
Chris L T521
Quote:

Originally Posted by SmallMan
a) Find the solutnion of the following differential equation:

(d^2y)/(dx^2)+2dy/dx-8y=0

b) Find the solutnion of the following differential equation:

(d^2y)/(dx^2)+2dy/dx-8y=14e^3x

With initial conditions: at x=0, y(0)=3 and (dy/dx)(0)=-4

My last problem, i tried to solve but every time i was getting other solutions. Can somebody tell me (especially) explain how to solve this ?

a) Note that the characteristic equation associated with the DE is $r^2+2r-8=0$. Thus, we see that $\left(r+4\right)\left(r-2\right)=0\implies r=-4$ or $r=2$.

Thus, our general solution is of the form $y=c_1e^{-4x}+c_2e^{2x}$.

Does this make sense?

b) First, consider the non-homogeneous equation (which is in (a) ). So, by the method of undetermined coefficients, the particular solution with take on the form $y_p=Ae^{3x}$.

Now, we see that

$\frac{\,dy_p}{\,dx}=3Ae^{3x}$

$\frac{\,d^2y_p}{\,dx^2}=9Ae^{3x}$

So if we substitute this into the original equation, it follows that

$\left(9Ae^{3x}\right)+2\left(3A\right)e^{3x}-8\left(Ae^{3x}\right)=14e^{3x}\implies7Ae^{3x}=14e ^{3x}$.

Now it's clear from here that $A=2$.

Therefore, the general solution is of the form $y=c_1e^{-4x}+c_2e^{2x}+2e^{3x}$.

Now apply the initial conditions $y\left(0\right)=3$, $y^{\prime}\left(0\right)=-4$

Applying the first condition gives us $3=c_1+c_2+2$

Differentiate our solution and apply the second initial condition to get

$-4=-4c_1+2c_2+2$

Now solve the system of equations to get $c_1$ and $c_2$.

Then you'll have you solution to the DE.

Does this make sense?
• Aug 19th 2009, 11:11 PM
SmallMan
Thanks a lot people :)