General Solution of a Differential Equation with Given Solution y=e^x

[Thumper88]

Homework Statement

Consider the differential equation:

(x^2 + x) \frac {d^2y} {dx^2} - (x^2 - 2) \frac {dy} {dx} - (x + 2)y = 0

Given that y = e^x is a solution, what is the general solution of this equation?

Homework Equations

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The Attempt at a Solution

I don't know where to start...please help :(

 

[gabbagabbahey]

Since you are already given one solution, this would seem to suggest using the method of reduction of order...so assume an ansatz y_2(x)=v(x)y_1(x)=v(x)e^x where v(x) is an unkown function of x...If you use the chain rule to find the first and second derivatives of y_2(x) (they will involve v(x) and its derivatives) and substitute them into your original DE, you can obtain a differential equation for v(x) which should be easier to solve. Then just solve that DE and discard any solutions that make y_2(x) linearly dependent on y_1(x).

 

[Thumper88]

So I have the derivatives (u = v(x)):

y_2 = u y_1 = ue^x

y_2' = uy_1' + y_1u' = ue^x + e^xu'

y_2'' = uy_1'' + 2u'y_1' + y_1u'' = ue^x + 2u'e^x + e^xu''

Putting these into the original DE...

(x^2 + x) \frac {d^2y} {dx^2} - (x^2 - 2) \frac {dy} {dx} - (x + 2)y = 0

(x^2 + x)(ue^x + 2u'e^x + e^xu'') - (x^2 - 2)(ue^x + e^xu') - (x + 2)(ue^x) = 0

Not sure where to go with it next...my teacher gave a hint that there will be integration of parts.

 

[Thumper88]

ok, well...to further things a little...

I distributed everything out and added like terms, leading to:

3u'x^2e^x + 2u'xe^x - 2u'e^x + u''x^2e^x + u''xe^x = 0

Using the variable transformation, w = u' and w' = u''

3wx^2e^x + 2wxe^x - 2we^x + w'x^2e^x + w'xe^x = 0

Still looks like a mess to me...?

 

[Thumper88]

yea, I'm at a halt here..really not sure what to do next.

 

[gabbagabbahey]

ok, well...to further things a little...

I distributed everything out and added like terms, leading to:

3u'x^2e^x + 2u'xe^x - 2u'e^x + u''x^2e^x + u''xe^x = 0

You might want to check your algebra again; I get:

u'x^2e^x + 2u'xe^x + 2u'e^x + u''x^2e^x + u''xe^x = 0

\Rightarrow e^x[u''(x^2+x) +u'(x^2+2x+2)] = 0

\Rightarrow u''(x^2+x) +u'(x^2+2x+2)=0

\Rightarrow w'(x^2+x) + w(x^2+2x+2)=0

which, when you write w'=\frac{dw}{dx}, looks like a separable DE for w to me...you know how to deal with those don't you?

 

[Thumper88]

I redid the algebra by hand and got exactly what you got.

Separating the variables

-\frac {1}{w} dw = \frac {(x^2+2x+2)}{(x^2+x)}dx

integrating both sides and solving for w I get:

w = \frac {((x+1) * e^-x)}{x^2} dx

So now u = integral of w dx

 

[gabbagabbahey]

I assume you mean: w=\frac{(x+1)e^{-x}}{x^2}?...if so, then good...now solve for u(x) using your definition of w(x): w(x)=u'(x)...then finally solve for y_2(x)=u(x)e^x...what do you get?

 

[Thumper88]

yea...i was editing the equation...still getting used to the code

 

[Thumper88]

u(x) = -4 e^{-x}

so y_2(x) = -4 e^{-x} * e^x = -4

correct?

 

[Thumper88]

so would the general solution just be:

y(x) = C_1 e^x + C_2
Where -4 is included in the constant C_2.

 

[gabbagabbahey]

u(x) = -4 e^{-x}

so y_2(x) = -4 e^{-x} * e^x = -4

correct?

nope, I get u(x) = \frac{e^{-x}}{x}

give or take a constant multiplier.

...why don't you show me your work for integrate wdx?

 

[Thumper88]

u(x) = \int w dx

u(x) = \int \frac {(x+1) e^-x}{x^2} dx = \int \frac {e^{-x}}{x} dx + \int \frac {e^{-x}}{x^2} dx

\int u dv = uv - \int v du

\int \frac {e^{-x}}{x} dx

u = \frac {1}{x} du = -\frac {1}{x^2}dx

dv = e^{-x} dx v = -e^{-x}

\int u v' = \frac {1}{x}(-e^{-x}) - \int -e^{-x} * - \frac {1}{x^2} dx = -2e^{-x}

Used same method for \int \frac e^{-x}}{x^2}dx with different u du values, and got same answer, adding together got the -4e^{-x}

 

[Thumper88]

arg...made some typos on my calculator...not getting those answers anymore.

 

[gabbagabbahey]

u(x) = \int w dx

u(x) = \int \frac {(x+1) e^-x}{x^2} dx = \int \frac {e^{-x}}{x} dx + \int \frac {e^{-x}}{x^2} dx

\int u dv = uv - \int v du

\int \frac {e^{-x}}{x} dx

u = \frac {1}{x} du = -\frac {1}{x^2}dx

dv = e^{-x} dx v = -e^{-x}

\int u v' = \frac {1}{x}(-e^{-x}) - \int -e^{-x} * - \frac {1}{x^2} dx = -2e^{-x}

Do you mean:

\int \frac {e^{-x}}{x} dx=\int uv'=\frac{-e^{-x}}{x}- \int \frac{e^{-x}}{x^2}dx

...if so, there is no need to evaluate the integral on the right, since you are adding this to your other term:

\Rightarrow u(x) = \int \frac {e^{-x}}{x} dx + \int \frac {e^{-x}}{x^2} dx=\left(\frac{-e^{-x}}{x}- \int \frac{e^{-x}}{x^2}dx \right)+ \int \frac {e^{-x}}{x^2} dx=\frac{-e^{-x}}{x}

...I'm having a hard time seeing where your 4e^-x came from, but this is the way you should have done it.

 

[Thumper88]

the -4e^-x came from using a minus sign instead of the negative key...

 

[gabbagabbahey]

the -4e^-x came from using a minus sign instead of the negative key...

oh, lol...that's why I don't like using calculators

 

[Thumper88]

been a while since I've done any integration by parts...

ok...so with
u(x) = - \frac {e^{-x}}{x}

y_2(x) = - \frac {e^{-x}}{x} * e^x = -\frac {1}{x}

Sooo...

y(x) = C_1 e^x + C_2 (-x^{-1})

 

[Thumper88]

that is correct, right? just checking, thanks for all the help!

 

[gabbagabbahey]

Looks good to me

...although your prof might prefer that you make sure y1 and y2 are linearly independent by checking the Wronskian.

 

[Thumper88]

I can do that...lol. Thanks a lot.