- #1

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I don't understand how to equate the coefficients and I was stuck after taking the integral of [P(x)y']'+[f(x)y]'=0 and got [P(x)y']+[f(x)y]=0

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- Thread starter physicsss
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- #1

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I don't understand how to equate the coefficients and I was stuck after taking the integral of [P(x)y']'+[f(x)y]'=0 and got [P(x)y']+[f(x)y]=0

- #2

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all right, let's start with the basics:

1. Py'' + Qy' + Ry = 0

is said to be exact if it can be written in the form

2. (My')' + (Ny)' = 0

Where P, Q, R, M, N, are functions only of x.

prime is derivative wrt x. and we have existence

of M'', P'', N', Q'. and untill I know better assume

none of the above is 0.

if we take the derivatives indicated in 2.

M'y' + My'' + N'y + Ny' = 0

or

My'' + (M' + N)y' + N'y = 0.

Py'' + Qy' + Ry = 0

now try equating the coefficients and generating

the necessary condition. [in this case, you need

to eliminate both M, and N.]

to solve (My')' +(Ny)' = 0

we integrate once to get My' + Ny = C

where C is an integration constant

we can put this in standard form

y' + (N/M)y = C/M

this is solvable by picking a suitable integrating factor....

[now, which conditions on P, Q, R, N, M were necessary?]

1. Py'' + Qy' + Ry = 0

is said to be exact if it can be written in the form

2. (My')' + (Ny)' = 0

Where P, Q, R, M, N, are functions only of x.

prime is derivative wrt x. and we have existence

of M'', P'', N', Q'. and untill I know better assume

none of the above is 0.

if we take the derivatives indicated in 2.

M'y' + My'' + N'y + Ny' = 0

or

My'' + (M' + N)y' + N'y = 0.

Py'' + Qy' + Ry = 0

now try equating the coefficients and generating

the necessary condition. [in this case, you need

to eliminate both M, and N.]

to solve (My')' +(Ny)' = 0

we integrate once to get My' + Ny = C

where C is an integration constant

we can put this in standard form

y' + (N/M)y = C/M

this is solvable by picking a suitable integrating factor....

[now, which conditions on P, Q, R, N, M were necessary?]

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- #3

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So [P(x)y']'+[f(x)y]'=0 -> P(x)y"+(P'(x)+f(x))y'+f '(x)y=0, and equating them I got:

P(x)=P(x),

Q(x)=P'(x)+f(x) -> f '(x)= -P"(x)+Q'(x)

R(x)= f '(x)

Then setting -P"(x)+Q'(x) = R(x) -> P"(x)-Q'(x)+R(x)=0

Will that suffice for showing the condition?

P(x)=P(x),

Q(x)=P'(x)+f(x) -> f '(x)= -P"(x)+Q'(x)

R(x)= f '(x)

Then setting -P"(x)+Q'(x) = R(x) -> P"(x)-Q'(x)+R(x)=0

Will that suffice for showing the condition?

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- #4

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I got it into this form: (y')'+(xy)'=0, integrated and got y'+(xy)=C. And I don't know where to go from here to find C.

- #5

saltydog

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Homework Helper

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qbert said:My'' + (M' + N)y' + N'y = 0.

Py'' + Qy' + Ry = 0

Thanks Qbert. That's very interesting. Can I offer some help and kindly correct me if I'm wrong because this is new to me too.

Physicsss, look carefully at those two equations Qbert gave us. Very carefully. Isn't P=M, R is something, but look at the middle part:

Isn't Q equal to the derivative of M (which is P) plus the antiderivative of N' which is R? You know, if R is equal to N' then N is equal to the antiderivative of R. So one can look at an equation:

[tex]Py^{''}+Qy^{'}+Ry=0[/tex]

and ask immediately: is the center function equal to the derivative of the first function plus the antiderivative of the third function? Is this correct Qbert?

Now, how about solving this equation which I'm working on too using this method:

[tex]x^2 y^{''}+(2x+\frac{1}{4} x^4)y^{'}+x^3y=0[/tex]

Any initial condtions, say y(0.1)=0 and y'(0.1)=1 (stay away from singularity at x=0) or whatever initial conditions are easy.

Edit: I mean you Physicsss, not you Qbert.

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- #6

HallsofIvy

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physicsss said:

I got it into this form: (y')'+(xy)'=0, integrated and got y'+(xy)=C. And I don't know where to go from here to find C.

Unless you have some additional conditions you can't- C is a "constant of integration". If you solve the first order equation y'+ xy= C, then you get another constant. That's what you would expect- when you solve a second order differential equation, you expect the general solution to involve two undetermined constants.

If the problem were y"+ xy'+ y= 0 with initial conditions: y(0)= 1, y'(0)= 0, then, taking x= 0 in y'+ xy= C, we get 0+ 0(1)= C or C= 0. That makes the equation y'+ xy= 0 which is easy (I cleverly chose y'(0)= 0 to get that!).

dy= -xy dx or dy/y= -xd so ln|y|= -x

At x= 0 that becomes y= 1= E e

- #7

saltydog

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Physicsss, where you at? Try and follow this one and then I have one for you to try if you want. The technique Qbert suggested is new to me so this is my first attempt at using such and applying it to an IVP:

For the equation:

[tex]Py^{''}+Qy^{'}+Ry=0[/tex]

If it's exact we can immediately write:

[tex](Py^{'})^{'}+(y\int R)^{'}=0[/tex]

Example:

[tex]x^2y^{''}+(2x+\frac{1}{4}x^4)y^{'}+x^3y=0;\quad y(0.1)=0,\quad y^{'}(0.1)=1[/tex]

Noting that the middle function is the sum of the derivative of the first function and the antiderivative of the last function, we can then say:

[tex](x^2y^{'})^{'}+(\frac{1}{4}x^4y)^{'}=0[/tex]

Taking antiderivatives:

[tex]\int (x^2y^{'})^{'}+\int(\frac{1}{4}x^4y)^{'}=\int 0[/tex]

Yielding:

[tex]x^2y^{'}+1/4x^4y=c_1[/tex]

Setting up this equation in the standard form to determine the integrating factor, we divide by [itex]x^2[/itex]

[tex]y^{'}+1/4x^2y=c_1x^{-2}[/tex]

The integrating factor is:

[tex]\sigma=Exp\left(\int 1/4 x^2dx\right)=e^{x^3/12}[/tex]

Multiplying the ODE by [itex]\sigma[/itex] converts the LHS to an exact differential:

[tex]d(e^{x^3/12}y)=c_1x^{-2}e^{x^3/12}[/tex]

Integrating:

[tex]\int_{\xi=0.1,\zeta=0}^{\xi=x,\zeta=y} d(e^{\xi^3/12}\zeta)=c_1\int_{0.1}^{x}t^{-2}e^{t^3/12}dt[/tex]

This yields:

[tex]e^{x^3/12}y(x)=c_1\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

or:

[tex]y(x)=c_1e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

So how do we find the constant? What about differentiating y(x) using Leibnitz's rule for the integral:

[tex]y^{'}(x)=c_1\left[e^{-x^3/12}\frac{d}{dx}\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)+\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)\frac{d}{dx}e^{-x^3/12}\right][/tex]

I know it's messy but try to follow it alright?

Now, we know from the IVP that the derivative at 0.1 is 1. Now look at the second sum. When we insert x=0.1 into the integral, that will be zero right? We're left with then:

[tex]y^{'}(x)=c_1e^{-x^3/12}\left(x^{-2}e^{x^3/12}\right)=\frac{c_1}{x^2}[/tex]

Solving:

[tex]y^{'}(0.1)=\frac{c_1}{(0.1)^2}=1[/tex]

We have [itex]c_1=0.01[/itex]

Thus the solution is:

[tex]y(x)=0.01e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

Now, there's nothing wrong with leaving it that way. Perfectly fine. You're comfortable with functions like Sin(x), Tan(x) right? What about a Sal(x)?

Let:

[tex]Sal(x)=\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

Then the solution is simply:

[tex]y(x)=0.01e^{-x^3/12}Sal(x)[/tex]

No different than using Sin(x) ok?

A plot is attached.

Now try this one:

[tex]6y^{''}+(\cos x) y^{'}-(\sin x) y=0; \quad y(0)=0,\quad y^{'}(0)=1[/tex]

I have no idea how it will turn out.

For the equation:

[tex]Py^{''}+Qy^{'}+Ry=0[/tex]

If it's exact we can immediately write:

[tex](Py^{'})^{'}+(y\int R)^{'}=0[/tex]

Example:

[tex]x^2y^{''}+(2x+\frac{1}{4}x^4)y^{'}+x^3y=0;\quad y(0.1)=0,\quad y^{'}(0.1)=1[/tex]

Noting that the middle function is the sum of the derivative of the first function and the antiderivative of the last function, we can then say:

[tex](x^2y^{'})^{'}+(\frac{1}{4}x^4y)^{'}=0[/tex]

Taking antiderivatives:

[tex]\int (x^2y^{'})^{'}+\int(\frac{1}{4}x^4y)^{'}=\int 0[/tex]

Yielding:

[tex]x^2y^{'}+1/4x^4y=c_1[/tex]

Setting up this equation in the standard form to determine the integrating factor, we divide by [itex]x^2[/itex]

[tex]y^{'}+1/4x^2y=c_1x^{-2}[/tex]

The integrating factor is:

[tex]\sigma=Exp\left(\int 1/4 x^2dx\right)=e^{x^3/12}[/tex]

Multiplying the ODE by [itex]\sigma[/itex] converts the LHS to an exact differential:

[tex]d(e^{x^3/12}y)=c_1x^{-2}e^{x^3/12}[/tex]

Integrating:

[tex]\int_{\xi=0.1,\zeta=0}^{\xi=x,\zeta=y} d(e^{\xi^3/12}\zeta)=c_1\int_{0.1}^{x}t^{-2}e^{t^3/12}dt[/tex]

This yields:

[tex]e^{x^3/12}y(x)=c_1\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

or:

[tex]y(x)=c_1e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

So how do we find the constant? What about differentiating y(x) using Leibnitz's rule for the integral:

[tex]y^{'}(x)=c_1\left[e^{-x^3/12}\frac{d}{dx}\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)+\left(\int_{0.1}^x t^{-2}e^{t^3/12}dt\right)\frac{d}{dx}e^{-x^3/12}\right][/tex]

I know it's messy but try to follow it alright?

Now, we know from the IVP that the derivative at 0.1 is 1. Now look at the second sum. When we insert x=0.1 into the integral, that will be zero right? We're left with then:

[tex]y^{'}(x)=c_1e^{-x^3/12}\left(x^{-2}e^{x^3/12}\right)=\frac{c_1}{x^2}[/tex]

Solving:

[tex]y^{'}(0.1)=\frac{c_1}{(0.1)^2}=1[/tex]

We have [itex]c_1=0.01[/itex]

Thus the solution is:

[tex]y(x)=0.01e^{-x^3/12}\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

Now, there's nothing wrong with leaving it that way. Perfectly fine. You're comfortable with functions like Sin(x), Tan(x) right? What about a Sal(x)?

Let:

[tex]Sal(x)=\int_{0.1}^x t^{-2}e^{t^3/12}dt[/tex]

Then the solution is simply:

[tex]y(x)=0.01e^{-x^3/12}Sal(x)[/tex]

No different than using Sin(x) ok?

A plot is attached.

Now try this one:

[tex]6y^{''}+(\cos x) y^{'}-(\sin x) y=0; \quad y(0)=0,\quad y^{'}(0)=1[/tex]

I have no idea how it will turn out.

Last edited:

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