Where have I gone wrong? (FP2) Watch

Chlorophile
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#1
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#1
Solve the equation (1-x^2)\frac{dy}{x}+xy=5x

\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

\frac{dy}{dx}\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}\frac{xy}{1-x^2}=\frac{1}{\sqrt{1-x^2}}\frac{5x}{1-x^2}

\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

\frac{y}{\sqrt{1-x^2}}=-\frac{3}{2}(1-x^2)^{\frac{5}{3}}+C

And from then on it's quite obviously going to go wrong. Where did I make a mistake/s? This is a complete trainwreck and my guess is that it was a basic error very early on, but I'm not sure.
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joostan
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#2
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#2
(Original post by Chlorophile)
Solve the equation (1-x^2)\frac{dy}{x}+xy=5x

\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

\frac{dy}{dx}\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}\frac{xy}{1-x^2}=\frac{1}{\sqrt{1-x^2}}\frac{5x}{1-x^2}

\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

\frac{y}{\sqrt{1-x^2}}=-\frac{3}{2}(1-x^2)^{\frac{5}{3}}+C

And from then on it's quite obviously going to go wrong. Where did I make a mistake/s? This is a complete trainwreck and my guess is that it was a basic error very early on, but I'm not sure.
Your integral: \frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx
Is not correct.
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jameswhughes
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#3
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#3
(Original post by Chlorophile)
Solve the equation (1-x^2)\frac{dy}{x}+xy=5x

\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

\frac{dy}{dx}\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}\frac{xy}{1-x^2}=\frac{1}{\sqrt{1-x^2}}\frac{5x}{1-x^2}

\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

\frac{y}{\sqrt{1-x^2}}=-\frac{3}{2}(1-x^2)^{\frac{5}{3}}+C

And from then on it's quite obviously going to go wrong. Where did I make a mistake/s? This is a complete trainwreck and my guess is that it was a basic error very early on, but I'm not sure.
Have a look at that last integral again, you've got \frac{3}{2} mixed up with \frac{2}{3}
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Chlorophile
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#4
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(Original post by joostan)
Your integral: \frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx
Is not correct.
I integrated the 5x(1-x^2)^1.5 part incorrectly?
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TeeEm
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#5
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#5
(Original post by Chlorophile)
Solve the equation (1-x^2)\frac{dy}{x}+xy=5x

\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

\frac{dy}{dx}\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}\frac{xy}{1-x^2}=\frac{1}{\sqrt{1-x^2}}\frac{5x}{1-x^2}

\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

\frac{y}{\sqrt{1-x^2}}=-\frac{3}{2}(1-x^2)^{\frac{5}{3}}+C

And from then on it's quite obviously going to go wrong. Where did I make a mistake/s? This is a complete trainwreck and my guess is that it was a basic error very early on, but I'm not sure.

apart from an obvious typo in the very last line, I cannot see at first glance anything wrong.

why do you think it is wrong?
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joostan
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#6
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#6
(Original post by Chlorophile)
I integrated the 5x(1-x^2)^1.5 part incorrectly?
Yup you might want to look at your arithmetic again.
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Chlorophile
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#7
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#7
Right, I realised what I did wrong. For some reason I thought 1/(1-x^2)^3/2 = (1-x^2)^2/3 (embarrassing). Thanks! (Virtual rep for all of you, I've run out for today)
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JerzyDudek
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#8
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Check your integration when finding the integrating factor. The integral you've written down doesn't evaluate to that.
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TeeEm
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#9
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(Original post by JerzyDudek)
Check your integration when finding the integrating factor. The integral you've written down doesn't evaluate to that.
how come?
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JerzyDudek
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#10
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#10
(Original post by TeeEm)
how come?
Oh pardon me, I've misread the notation.
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TeeEm
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#11
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#11
(Original post by JerzyDudek)
Oh pardon me, I've misread the notation.
no worries
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Rkai01
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#12
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#12
A bit random but would anyone say dropping M1 for FP2 is stupid as I dislike Mechanics?Also pure modules are helpful for someone like me as Im doing Economics and a lot of Further methods are used I believe whereas I would almost never use any M1 methods in econ.
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TeeEm
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#13
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(Original post by Rkai01)
A bit random but would anyone say dropping M1 for FP2 is stupid as I dislike Mechanics?Also pure modules are helpful for someone like me as Im doing Economics and a lot of Further methods are used I believe whereas I would almost never use any M1 methods in econ.
Maybe you should start your own thread.
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