2nd order DE's with constant coefficients.

**hejmh** · January 8th 2011, 03:31 PM

1- How can I solve this equation? y''+4y'=0

2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:
e^(3x) and xe^(3x)

Thanks for your help.

**Ackbeet** · January 8th 2011, 03:32 PM

What ideas have you had so far?

**dwsmith** · January 8th 2011, 03:38 PM

Originally Posted by hejmh

1- How can I solve this equation? y''+4y'=0

2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:
e^(3x) and xe^(3x)

Thanks for your help.

$y''+4y'=0\Rightarrow m^2+4m=0$

Do you know how to use this form?

**Prove It** · January 8th 2011, 03:49 PM

For part 2, you're told $\displaystyle y = e^{3x}$ is a solution.

So $\displaystyle y' = 3e^{3x}$ and $\displaystyle y'' = 9e^{3x}$ .

Substitute into the DE to get $\displaystyle 9e^{3x} + 3a\,e^{3x} + b\,e^{3x} = 0$ .

Now do the same with the other solution $\displaystyle y = x\,e^{3x}$ to get another equation that involves $\displaystyle a$ and $\displaystyle b$ . Then you can solve the two equations simultaneously for $\displaystyle a$ and $\displaystyle b$ .

**dwsmith** · January 8th 2011, 04:00 PM

Originally Posted by dwsmith

$y''+4y'=0\Rightarrow m^2+4m=0$

Do you know how to use this form?

From this point, you factor and solve for m.

$m(m+4)=0\Rightarrow m=0,-4$

$y=C_1e^{m_1x}+C_2e^{m_2x}$

$y=C_1e^{0*x}+C_2e^{-4x}=C_1+C_2e^{-4x}$

**hejmh** · January 8th 2011, 04:02 PM

Yes. thanks

**HallsofIvy** · January 9th 2011, 07:19 AM

Originally Posted by hejmh

1- How can I solve this equation? y''+4y'=0

Here's another way to do this. Since there is no "y" (undifferentiated) let u= y' so the equation becomes u'+ 4u= 0. That is the same as du/dx= -4u which is a separable, first order differential equation. You can write it as du/u= -4 dx and integrate both sides: ln(u)= -4x+ c which leads to $u(x)= y'(x)= e^{-4x+ c}= Ce^{-4x}$ where $C= e^{c}$ . Now find y by integrating $y'= Ce^{-4x}$ : $y(x)= \frac{C}{-4}e^{4x}+ B$ which we can also write as $y(x)= Ae^{-4x}+ B$ where now $A= \frac{C}{-4}= \frac{e^c}{-4}$ .

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