Stuck on these differential equations

[James10492]

Hi there, I have just come across a couple of differentials I cannot solve. I have applied the methods I usually use for solving differential equations but frustratingly my solutions do not match the ones in my book. Please could someone point out what I am not doing right?

Problem one:

given that y = 6 and x = .5, find the particular solution:

[math](1-x^2)\frac{dy}{dx} = xy + y[/math]
[math](1-x^2)\frac{dy}{dx} = y(x+1)[/math]
[math]\frac{dy}{y} = \frac{(1+x)}{(1+x)(1-x)} dx[/math]
[math]\int\frac{dy}{y} = \int\frac{1}{(1-x)}[/math]
[math]\ln{y} = \ln{(1-x)} + \ln{k}[/math]
[math]y = k(1-x)[/math]
[math]6 = k(1-.5) \\ 6= .5k \\ k = 12[/math]
so the particular solution is 12(1-x), right?


Problem two:

find the particular solution for y=2, x =0:

[math](1+x^2)\frac{dy}{dx} = x - xy^2 \\ (1+x^2)\frac{dy}{dx} = x(1 - y^2) \\ \frac{dy}{1-y^2} = \frac{x}{1+x^2}dx \\[/math]
writing the left-hand side as partial fractions:

[math]\int\frac{1}{2(1+y)} + \frac{1}{2(1-y)} dy = \int\frac{x}{1+x^2} dx \\ \frac{1}{2} \ln{(1+y)} - \frac{1}{2}\ln(1-y) = \frac{1}{2}\ln(1+x^2) + C \\ \ln{(1+y)} - \ln(1-y) = \ln(1+x^2) + \ln {k}\\ \frac{(1+y)}{(1-y)} = k(1+x^2) \\ -\frac{3}{2} = k[/math]
I think something is probably wrong because now it is difficult to write y = f(x)

 

[topsquark]

Hi there, I have just come across a couple of differentials I cannot solve. I have applied the methods I usually use for solving differential equations but frustratingly my solutions do not match the ones in my book. Please could someone point out what I am not doing right?

Problem one:

given that y = 6 and x = .5, find the particular solution:

[math](1-x^2)\frac{dy}{dx} = xy + y[/math]
[math](1-x^2)\frac{dy}{dx} = y(x+1)[/math]
[math]\frac{dy}{y} = \frac{(1+x)}{(1+x)(1-x)} dx[/math]
[math]\int\frac{dy}{y} = \int\frac{1}{(1-x)}[/math]
[math]\ln{y} = \ln{(1-x)} + \ln{k}[/math]
[math]y = k(1-x)[/math]
[math]6 = k(1-.5) \\ 6= .5k \\ k = 12[/math]
so the particular solution is 12(1-x), right?


Problem two:

find the particular solution for y=2, x =0:

[math](1+x^2)\frac{dy}{dx} = x - xy^2 \\ (1+x^2)\frac{dy}{dx} = x(1 - y^2) \\ \frac{dy}{1-y^2} = \frac{x}{1+x^2}dx \\[/math]
writing the left-hand side as partial fractions:

[math]\int\frac{1}{2(1+y)} + \frac{1}{2(1-y)} dy = \int\frac{x}{1+x^2} dx \\ \frac{1}{2} \ln{(1+y)} - \frac{1}{2}\ln(1-y) = \frac{1}{2}\ln(1+x^2) + C \\ \ln{(1+y)} - \ln(1-y) = \ln(1+x^2) + \ln {k}\\ \frac{(1+y)}{(1-y)} = k(1+x^2) \\ -\frac{3}{2} = k[/math]
I think something is probably wrong because now it is difficult to write y = f(x)

What are your book answers? I see no problem with your work.

Go one little step to another:
[imath]\dfrac{1 - y}{1 + y} =- \dfrac{3}{2} (1 + x^2)[/imath]

[imath]1 - y = - \dfrac{3}{2} (1 + x^2) (1 + y)[/imath]

[imath]1 - y = - \dfrac{3}{2} (1 + x^2) - \dfrac{3}{2} (1 + x^2) y[/imath]

Now combine like terms, etc.

-Dan

 

[JeffM]

Hi there, I have just come across a couple of differentials I cannot solve. I have applied the methods I usually use for solving differential equations but frustratingly my solutions do not match the ones in my book. Please could someone point out what I am not doing right?

Problem one:

given that y = 6 and x = .5, find the particular solution:

[math](1-x^2)\frac{dy}{dx} = xy + y[/math]
[math](1-x^2)\frac{dy}{dx} = y(x+1)[/math]
[math]\frac{dy}{y} = \frac{(1+x)}{(1+x)(1-x)} dx[/math]
[math]\int\frac{dy}{y} = \int\frac{1}{(1-x)}[/math]
[math]\ln{y} = \ln{(1-x)} + \ln{k}[/math]
[math]y = k(1-x)[/math]
[math]6 = k(1-.5) \\ 6= .5k \\ k = 12[/math]
so the particular solution is 12(1-x), right?

On problem 1, I am with you up to [imath]\dfrac{dy}{y} = \dfrac{dx}{1-x}. [/imath]

I find I avoid many silly errors by using u-substitutions and paying very careful attention to differentials.

[math]u = (1 - x) \implies du = - dx \implies dx = - du \implies \dfrac{dy}{y} = \dfrac{dx}{1 - x} = - \dfrac{du}{u} \implies \\ \int \dfrac{dy}{y} = - \int \dfrac{du}{u} \implies ln(y) = - ln(u) + ln(k) = ln \left ( \dfrac{k}{u} \right ) \implies \\ y = \dfrac{k}{u} = \dfrac{k}{1 - x}.\\ \text {But } f(0.5) = 6 \implies 6 = \dfrac{k}{1 - 0.5} \implies k = 6(1 - 0.5) = 3.[/math]
Does that agree with your book?

I have not looked at problem 2. The rules say one problem per thread.

 

[James10492]

On problem 1, I am with you up to [imath]\dfrac{dy}{y} = \dfrac{dx}{1-x}. [/imath]

I find I avoid many silly errors by using u-substitutions and paying very careful attention to differentials.

[math]u = (1 - x) \implies du = - dx \implies dx = - du \implies \dfrac{dy}{y} = \dfrac{dx}{1 - x} = - \dfrac{du}{u} \implies \\ \int \dfrac{dy}{y} = - \int \dfrac{du}{u} \implies ln(y) = - ln(u) + ln(k) = ln \left ( \dfrac{k}{u} \right ) \implies \\ y = \dfrac{k}{u} = \dfrac{k}{1 - x}.\\ \text {But } f(0.5) = 6 \implies 6 = \dfrac{k}{1 - 0.5} \implies k = 6(1 - 0.5) = 3.[/math]
Does that agree with your book?

I have not looked at problem 2. The rules say one problem per thread.

Thank you JeffM I had not tried that sorry I forgot about that rule

 

[Steven G]

[math] \frac{(1+y)}{(1-y)} = k(1+x^2)[/math]
I think something is probably wrong because now it is difficult to write y = f(x)

It is only difficult to write y=f(x) probably because you haven't tried. It is actually quite to easy to do so. If you are having trouble doing that then please post your work and you'll receive excellent help.

 

[James10492]

It is only difficult to write y=f(x) probably because you haven't tried. It is actually quite to easy to do so. If you are having trouble doing that then please post your work and you'll receive excellent help.

It looked like it might have been more difficult than it should have been, which was why I wanted to check my method ( you are right, I just didn't try hard enough).

 

[JeffM]

As Steven says, it is really not hard to come up with f(x) if you remember some things that you learned in first-year algebra. It is less messy if you do that BEFORE solving for k. You are solving for one variable in terms of one or more other variables.

Expand parentheticals: [imath]\dfrac{1 + y}{1 - y} = k(1 + x^2) = k + kx^2[/imath].

Clear fractions: [imath]1 + y = k + kx^2 - ky - kx^2y[/imath].

Isolate variable of interest on LHS: [imath]y + ky +kx^2y = kx^2 + k - 1[/imath].

Use distribution property: [imath]y(kx^2 + k + 1) = kx^2 + k - 1.[/imath]

Divide: [imath]y = \dfrac{kx^2 + k - 1}{kx^2 + k + 1}[/imath].

Of course to find k, you must solve for k rather than y. But it is all first-year algebra. The hard part of calculus is remembering all your algebra.