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Where have I gone wrong? (FP2)

Solve the equation (1−x2)dyx+xy=5x(1-x^2)\frac{dy}{x}+xy=5x

dydx+xy1−x2=5x1−x2\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e∫x1−x2 dx=11−x2e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

dydx11−x2+11−x2xy1−x2=11−x25x1−x2\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}

d(y1−x2)dx=5x(1−x2)32\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

y1−x2=∫5x(1−x2)32 dx\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

y1−x2=−32(1−x2)53+C\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.
(edited 9 years ago)
Reply 1
Avatar for joostan
joostan
13
Original post by Chlorophile
Solve the equation (1−x2)dyx+xy=5x(1-x^2)\frac{dy}{x}+xy=5x

dydx+xy1−x2=5x1−x2\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e∫x1−x2 dx=11−x2e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

dydx11−x2+11−x2xy1−x2=11−x25x1−x2\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}

d(y1−x2)dx=5x(1−x2)32\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

y1−x2=∫5x(1−x2)32 dx\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

y1−x2=−32(1−x2)53+C\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: y1−x2=∫5x(1−x2)32 dx\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx
Is not correct.
Original post by Chlorophile
Solve the equation (1−x2)dyx+xy=5x(1-x^2)\frac{dy}{x}+xy=5x

dydx+xy1−x2=5x1−x2\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e∫x1−x2 dx=11−x2e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

dydx11−x2+11−x2xy1−x2=11−x25x1−x2\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}

d(y1−x2)dx=5x(1−x2)32\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

y1−x2=∫5x(1−x2)32 dx\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

y1−x2=−32(1−x2)53+C\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 32\frac{3}{2} mixed up with 23\frac{2}{3}
Original post by joostan
Your integral: y1−x2=∫5x(1−x2)32 dx\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?
Reply 4
Avatar for TeeEm
TeeEm
19
Original post by Chlorophile
Solve the equation (1−x2)dyx+xy=5x(1-x^2)\frac{dy}{x}+xy=5x

dydx+xy1−x2=5x1−x2\frac{dy}{dx}+\frac{xy}{1-x^2}=\frac{5x}{1-x^2}

Integrating factor = e∫x1−x2 dx=11−x2e^{\displaystyle\int \frac{x}{1-x^2}\ dx} = \frac{1}{\sqrt{1-x^2}}

dydx11−x2+11−x2xy1−x2=11−x25x1−x2\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}

d(y1−x2)dx=5x(1−x2)32\frac{d(\frac{y}{\sqrt{1-x^2}})}{dx}=\frac{5x}{({1-x^2})^{\frac{3}{2}}}

y1−x2=∫5x(1−x2)32 dx\frac{y}{\sqrt{1-x^2}}=\displaystyle\int \frac{5x}{({1-x^2})^{\frac{3}{2}}}\ dx

y1−x2=−32(1−x2)53+C\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?
Reply 5
Avatar for joostan
joostan
13
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.
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)
Check your integration when finding the integrating factor. The integral you've written down doesn't evaluate to that.
Reply 8
Avatar for TeeEm
TeeEm
19
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?
Original post by TeeEm
how come?


Oh pardon me, I've misread the notation.
Reply 10
Avatar for TeeEm
TeeEm
19
Original post by JerzyDudek
Oh pardon me, I've misread the notation.


no worries
Reply 11
Avatar for Rkai01
Rkai01
6
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.
Reply 12
Avatar for TeeEm
TeeEm
19
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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