Homogenous Differential Equation Solution

[_N3WTON_]

Homework Statement

Solve the differential equation:
\frac{dy}{dx} = x + y

Homework Equations

Integrating factor equation

The Attempt at a Solution

Ok, I am aware that in order to solve this equation, I need to make a substitution:
v = x + y
However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.

 

[Ray Vickson]

Homework Statement

Solve the differential equation:
\frac{dy}{dx} = x + y

Homework Equations

Integrating factor equation

The Attempt at a Solution

Ok, I am aware that in order to solve this equation, I need to make a substitution:
v = x + y
However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.

The substitution $v = x+y$ leads to another inhomogeneous DE, but simpler than the original. Have you tried it?

I cannot believe that your textbook does not cover the methods for solving
\frac{dy}{dx} + f(x) y = g(x)
You say it does not cover this technique thoroughly, but surely it covers it at least a little bit. You need to show your work, and explain where/why you are stuck.

 

[LCKurtz]

@N3WTON: Alternatively, doesn't your text discuss constant coefficient homogeneous and non-homogeneous equations and the method of undetermined coefficients for NH equations?

 

[ehild]

Homework Statement

Solve the differential equation:
\frac{dy}{dx} = x + y

The Attempt at a Solution

Ok, I am aware that in order to solve this equation, I need to make a substitution:
v = x + y
However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.

With the substitution, y=v-x, y'=v'-1 and the differential equation becomes v'-1=v, that is, v'=v+1. Is it not a separable equation?

ehild

 

[_N3WTON_]

@N3WTON: Alternatively, doesn't your text discuss constant coefficient homogeneous and non-homogeneous equations and the method of undetermined coefficients for NH equations?

It does discuss constant coefficient homogenous and non-homogenous equations, however; no mention of the method of undetermined coefficients for NH equations. Perhaps we have just not reached that point in the text yet

 

[_N3WTON_]

The substitution $v = x+y$ leads to another inhomogeneous DE, but simpler than the original. Have you tried it?

I cannot believe that your textbook does not cover the methods for solving
\frac{dy}{dx} + f(x) y = g(x)
You say it does not cover this technique thoroughly, but surely it covers it at least a little bit. You need to show your work, and explain where/why you are stuck.

I have tried it but I don't understand what I need to do. The equation you listed could be solved using the integrating factor method, no? I'm confused about how that relates to the equation I listed...

 

[_N3WTON_]

With the substitution, y=v-x, y'=v'-1 and the differential equation becomes v'-1=v, that is, v'=v+1. Is it not a separable equation?

ehild

ok...so picking up from there I did:
\int\frac{v'}{v+1} = \int 0
ln(v+1) = C
v + 1 = e^c
v = e^{c} - 1
x + y = e^{c} - 1
Here I am stuck because the solution in the book is:
y = 2e^x - x - 1
Also the book notes that this problem can be "solved quite easily" using the integrating factor method, but I do not see it...

 

[pasmith]

ok...so picking up from there I did:
\int\frac{v'}{v+1} = \int 0

This should be \int \frac{v'}{v +1}\,dx = \int 1\,dx.

ln(v+1) = C
v + 1 = e^c
v = e^{c} - 1
x + y = e^{c} - 1
Here I am stuck because the solution in the book is:
y = 2e^x - x - 1

You appear to have omitted an initial condition (perhaps y(0) = 1) from your problem; otherwise '2' should instead be an arbitrary constant.

 

[ehild]

ok...so picking up from there I did:
\int\frac{v'}{v+1} = \int 0

What is v+1 divided by (v+1)??
ehild

 

[_N3WTON_]

This should be \int \frac{v'}{v +1}\,dx = \int 1\,dx.

nvm

 

[_N3WTON_]

thanks, but can you explain why that is correct?
before I had:
v' = v+1
\frac{v'}{v+1} = 0 <br /> I am not seeing where the dx comes from...

 

[Orodruin]

∫v′v+1,dv=∫0

You seem to have assumed that $v'/(v+1) = 0$. This is not the case so I suggest you recheck that algebra.

Another option is to find a particular solution (there is a fairly obvious one) and then add the solution to the homogeneous equation to that (which works since the ODE is linear).

 

[_N3WTON_]

This should be \int \frac{v'}{v +1}\,dx = \int 1\,dx.
You appear to have omitted an initial condition (perhaps y(0) = 1) from your problem; otherwise '2' should instead be an arbitrary constant.

you're right, I do apologize for that...I'll make the edit now

 

[_N3WTON_]

What is v+1 divided by (v+1)??
ehild

ok, I see the mistake now...thank you

 

[pasmith]

thanks, but can you explain why that is correct?
before I had:
v' = v+1
\frac{v'}{v+1} = 0
I am not seeing where the dx comes from...

You are integrating both sides with respect to x. It is crucial to state what variable you are integrating with respect to, because shortly you will use substitution to change \int \frac{v'}{v +1}\,dx into an integral with respect to v.

 

[_N3WTON_]

You are integrating both sides with respect to x. It is crucial to state what variable you are integrating with respect to, because shortly you will use substitution to change \int \frac{v'}{v +1}\,dx into an integral with respect to v.

thank you, so here is my second attempt:
\int\frac{v'}{v+1} dx = \int dx
ln(v+1) = x + C
v+1 = e^x + C
x + y = e^x + C

 

[ehild]

thank you, so here is my second attempt:
\int\frac{v'}{v+1} dx = \int dx
ln(v+1) = x + C
v+1 = e^x + C

the last line is wrong. It should be v+1 = e^{x + C} instead.

 

[_N3WTON_]

the last line is wrong. It should be v+1 = e^{x + C} instead.

thank you...

 

[Ray Vickson]

I have tried it but I don't understand what I need to do. The equation you listed could be solved using the integrating factor method, no? I'm confused about how that relates to the equation I listed...

In your example, is it not the case that $f(x) = -1$ and $g(x) = x$?

 

[ehild]

the last line is wrong. It should be v+1 = e^{x + C} instead.

And v+1 = e^{x + C} = Ae^x . The constants C or A depend on the initial condition.

ehild

 

[_N3WTON_]

And v+1 = e^{x + C} = Ae^x . The constants C or A depend on the initial condition.

ehild

ok, the initial condition was f(0) = 1...you arrived at that answer just by bringing the constant C in front of the exponential?

 

[_N3WTON_]

In your example, is it not the case that $f(x) = -1$ and $g(x) = x$?

ok so if i were to use the integrating factor method I would set:
p(x) = -1 ?
I'm still a bit confused about how I could do this problem using integrating factor method...I understand how to use separation of variables..

 

[Ray Vickson]

ok so if i were to use the integrating factor method I would set:
p(x) = -1 ?
I'm still a bit confused about how I could do this problem using integrating factor method...I understand how to use separation of variables..

There is a standard formula; just apply it to your problem.

 

[ehild]

ok, the initial condition was f(0) = 1...you arrived at that answer just by bringing the constant C in front of the exponential?

You know that $e^{{x+C}}=e^C e^x$, but e^C is also a constant, I denoted it by A. So v+1=y+x+1=Ae^x.. Applying the condition y(0)=1, 1+1=A.

 

[Orodruin]

Just for completeness, since we have a linear ODE with constant coefficients and a polynomial inhomogeneity, we could simply have assumed a particular solution of the form $y_p = ax + b$ and arrived at $a = b = -1$ and thus $y_p = -x-1$, $y'_p = -1$. We then write $y = y_p + y_h$ where $y_h$ satisfies the homogeneous ODE $y'_h = y_h$, which I believe anyone familiar with differential equations can solve ... This approach is what I was trying to hint at in post #12.

 

[HallsofIvy]

It is better NOT to memorize formulas (like that for the integrating factor of a linear differential equation) but to understand the basic concept and work from there. Here, the differential equation is dy/dx= x+ y which is the same as dy/dx- y= x. An "integrating factor" for this equation is a a function u(x) such that multiplying by it makes the left side a "complete derivative". That is, so that udy/dx- uy= d(uy)/dx. Using the product rule on the right, u dy/dx- uy= u dy/dx+ u'y. We can cancel the "u dy/dx" terms leaving -uy= u'y or u'= -u. Integrating that gives u(x)= Ce^{-x}.

Since we only need one, take C= 1 to get u(x)= e^{-x}. Multiplying both sides of the equation by that, we have e^{-x}dy/dx- e^{-x}y= d(e^{-x}y)/dx= xe^{-x}. Integrate both sides of that with respect to x.

 

[_N3WTON_]

You know that $e^{{x+C}}=e^C e^x$, but e^C is also a constant, I denoted it by A. So v+1=y+x+1=Ae^x.. Applying the condition y(0)=1, 1+1=A.

Thank you...your insight has been very helpful to me :)

 

[_N3WTON_]

Just for completeness, since we have a linear ODE with constant coefficients and a polynomial inhomogeneity, we could simply have assumed a particular solution of the form $y_p = ax + b$ and arrived at $a = b = -1$ and thus $y_p = -x-1$, $y'_p = -1$. We then write $y = y_p + y_h$ where $y_h$ satisfies the homogeneous ODE $y'_h = y_h$, which I believe anyone familiar with differential equations can solve ... This approach is what I was trying to hint at in post #12.

I appreciate your help, and I understand your solution. However, that is not a technique that we have covered in my DiffEq class so I don't know if I'd be able to use it on my upcoming exam...

 

[_N3WTON_]

It is better NOT to memorize formulas (like that for the integrating factor of a linear differential equation) but to understand the basic concept and work from there. Here, the differential equation is dy/dx= x+ y which is the same as dy/dx- y= x. An "integrating factor" for this equation is a a function u(x) such that multiplying by it makes the left side a "complete derivative". That is, so that udy/dx- uy= d(uy)/dx. Using the product rule on the right, u dy/dx- uy= u dy/dx+ u'y. We can cancel the "u dy/dx" terms leaving -uy= u'y or u'= -u. Integrating that gives u(x)= Ce^{-x}.

Since we only need one, take C= 1 to get u(x)= e^{-x}. Multiplying both sides of the equation by that, we have e^{-x}dy/dx- e^{-x}y= d(e^{-x}y)/dx= xe^{-x}. Integrate both sides of that with respect to x.

thank you, that actually helped me to better understand what I'm actually doing when I apply the integrating factor method...

 

[Orodruin]

I appreciate your help, and I understand your solution. However, that is not a technique that we have covered in my DiffEq class so I don't know if I'd be able to use it on my upcoming exam...

I think that the approach itself is not very different from the variable substitution that you started with (just writing $v = y + x + 1$ rather than $v = y+x$ - you could also see it as making yet another substitution from $v' = v + 1$ to $u = v+1$, which also gives $u' = u$). Regardless, only a very ignorant teacher would deduct points for you using a working method unless explicitly asking for something to be solved with a particular method. But of course this is completely up to you.

 

[_N3WTON_]

I think that the approach itself is not very different from the variable substitution that you started with (just writing $v = y + x + 1$ rather than $v = y+x$ - you could also see it as making yet another substitution from $v' = v + 1$ to $u = v+1$, which also gives $u' = u$). Regardless, only a very ignorant teacher would deduct points for you using a working method unless explicitly asking for something to be solved with a particular method. But of course this is completely up to you.

thats true...to my knowledge the test doesn't explicitly ask for a certain method so thanks for help :)

 

[ehild]

Thank you...your insight has been very helpful to me :)

You are welcome:)

ehild