# Solve 2 second order linear ODEs d^2y/dx^2 + 16(dy/dx) + 55y = 0, d^2y/dx^2 + 22(dy/dx) + 85y = 0, s.t. y(0) = 14

#### SerClaudio

##### New member

Solve the two differential equations:

. . . . .$$\displaystyle \dfrac{d^2y}{dx^2}\, +\, 16\, \dfrac{dy}{dx}\, +\, 55y\, =\, 0$$

. . . . .$$\displaystyle \dfrac{d^2y}{dx^2}\, +\, 22\, \dfrac{dy}{dx}\, +\, 85y\, =\, 0$$

...subject to $$\displaystyle y(0)\, =\, 14.$$

Thank you

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#### MarkFL

##### Super Moderator
Staff member
Can you begin by stating the characteristic/auxiliary equation for each ODE?

#### SerClaudio

##### New member
They will be respectively

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#### MarkFL

##### Super Moderator
Staff member
Yes, those are the general two-parameter solutions to the given ODEs. We see there is one solution that will satisfy both...can you identify it?

#### topsquark

##### Full Member
The two are being solved simultaneously, correct? As another approach I'd subtract the two equations. Then the second derivative drops out and you are left with a single linear equation.

-Dan

#### SerClaudio

##### New member
The two are being solved simultaneously, correct? As another approach I'd subtract the two equations. Then the second derivative drops out and you are left with a single linear equation.

-Dan
I tried this method and reckon it is the right approach to solve the problem. However, the linear expression I obtained is y=-5x+14 and when I put this back into either equation it does not add up to zero. To be a general solution it should work, but it doesnt?

#### SerClaudio

##### New member
Yes, those are the general two-parameter solutions to the given ODEs. We see there is one solution that will satisfy both...can you identify it?
I can use substitution and obtain an expression with both auxiliary equations but because the condition given is only one we would have 2 constants... it does not seem right

#### topsquark

##### Full Member
Yes, I realized last night that there is some kind of flaw in my reasoning. For example, MarkFL's method would immediately show if there is no solution, but mine will come up with a false one. I really have no idea why my method is flawed and won't always work.

Stick with MarkFL's method.

-Dan

#### SerClaudio

##### New member
Yes, I realized last night that there is some kind of flaw in my reasoning. For example, MarkFL's method would immediately show if there is no solution, but mine will come up with a false one. I really have no idea why my method is flawed and won't always work.

Stick with MarkFL's method.

-Dan
Your method and mine are right.. I simply made an implicit integration mistake. The general solution is y=Ae^(-5x) and the specific one is 14e^(-5x). Thank you anyway for your time.

#### MarkFL

##### Super Moderator
Staff member
I'm not seeing the flaw with Dan's method, and last night I thought it was a very efficient means to get where we want to be.. Subtracting the first ODE from the second (essentially applying superposition) leads to:

$$\displaystyle \d{y}{x}+5y=0$$

As this has a linear auxiliary equation with one root, this leads us to the general solution:

$$\displaystyle y(x)=c_1e^{-5x}$$

And this is the solution that satisfies both original 2nd order ODEs. Then to find the particular solution satisfying the given initial value:

$$\displaystyle y(0)=c_1=14\implies y(x)=14e^{-5x}$$

#### topsquark

##### Full Member
It certainly works in this instance, but suppose we have
$$\displaystyle \dfrac{d^2y}{dx^2} - 5 \dfrac{dy}{dx} + 6y = 0$$
and
$$\displaystyle \dfrac{d^2y}{dx^2} + 4 \dfrac{dy}{dx} + 3y = 0$$

My method gives $$\displaystyle y = E e^{-x/3}$$, but if you solve each equation separately you find that
$$\displaystyle y = Ae^{3x} + Be^{2x}$$ for the first equation
and
$$\displaystyle y = Ce^{-3x} + De^{-x}$$

There is no solution to this system but my method comes up with one anyway. I don't know how to test for the existence of a solution.

-Dan

#### MarkFL

##### Super Moderator
Staff member
Ah, okay I see what you mean. Thanks for that added information.