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Stuck trying to integrate a differential equation

  • Thread starter dooogle
  • Start date
  • #1
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Homework Statement



dy/dx=a^2/((x+y)^2)

where a is a constant

express answer in the form x=f(y)

Homework Equations


The Attempt at a Solution



let u=x+y

du/dx=1+dy/dx

du/dx=1+(a^2)/(u^2)

int(du/(((a^2)/(u^2))+1))=int(a^2 dx)

after integration substituting back in for u gives:

(-1/(2a))*arctan(a/((x+y)^2))=(a^2(x+c))/((x+y)^3)

i dont know how to rearrange into the form f(y)=x

please could you let me know if i have made mistakes and if not could u tell me how to rearrange arctan(a/(x+y)^2) into a form where x and y are separable

thanks for your time

dooogle
 

Answers and Replies

  • #2
188
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If I've interpreted your equations correctly (try using latex code, it is a lot simpler to read), it seems to me that you've done an error when simplifying the expression that is to be integrated. I get something like

[tex]\frac{du}{dx}=1+ \frac{a^2}{u^2}[/tex]

[tex]\Leftrightarrow \frac{du}{dx}=\frac{u^2 + a^2}{u^2} [/tex]

[tex]\Rightarrow \int \frac{u^2 du}{u^2 + a^2} = \int dx [/tex]

This integrand doesn't seem to agree with yours.
 
  • #3
21
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cheers for the response

ive tried integrating:

u^2/((u^2)+(a^2))du

by parts but the answer i get is not the same as what i get when i use wolfram online integrator

is it correct to try integration by parts? where the equation is

int(f(u)g'(u))=f(u)g(u)-int(f'(u)g(u)du)

taking f(u)=1/((u^2)+(a^2)) g'(u)=u^2

so f'(u)=-2u/((u^2)+(a^2))^2 g(u)=(u^3)/3

so integral equals

(u^3)/3((u^2)+(a^2))-int((u^3)/3*-2u/((u^2)+(a^2))^2))

=(u^3)/3((u^2)+(a^2))-int((-2u^4/(3((u^2)+(a^2))^2)))

which does not seem to work

i apologise for not using latex i tried to but it didnt work

thank you for your time any help would be much appreciated

cheers


dooogle
 
  • #4
Char. Limit
Gold Member
1,204
13
Try using the opposite order. Differentiate u2 and integrate 1/(u2+a2).
 
  • #5
21
0
ok so i set f(u)=u^2

f'(u)=2u

g'(u)=1/(u^2+a^2)

g(u)=ArcTan[u/a]/a

applying integration by parts formula gives

(u^2)*ArcTan[u/a]/a-int(2uArcTan[u/a]/a du)

which gives a*arctan(u/a)-x=int a^2 dx

arctan(u/a)=ax+c

so (u/a)=tan(ax) +tan(c)

subbing back in for x+y gives

x+y=a*tan(ax)+a*tan(c)

x=a*tan(ax)+a*tan(c)-y

please could u let me know if i have integrated incorrectly

or if i have made a mistake elsewhere

thanks very much for your time

dooogle
 
  • #6
Char. Limit
Gold Member
1,204
13
You integrated by parts incorrectly. You should get 2u/(u2+a2) as the second integral, not what you have.
 
  • #7
21
0
i have attempted it with a different method since the last method did not seem to supply the right answer please could someone tell me where i am going wrong with this new method:

dy/dx=(a^2)/((x+y)^2)

where a^2 is a constant and the answer must be expressed in the form x=f(y)

dy/dx=(a^2)/((x^2+2*x*y+y^2)

let u=y/x

dy/dx=(a^2)/((x^2+2u*x^2+u^2*x^2)

dy/dx=(a^2)/(x^2(1+2u+u^2)

du/dy=(du/dx)*(dx/dy)

du/dx=-y/(x^2)

dx/dy=(x^2(1+2u+u^2)/(a^2)

so du/dy=-y*((1+2u+u^2)/(a^2))

int(1/((u+1)^2))du= -1/a^2*int(y)dy

-1/(u+1)= -y^2/(2*a^2)+[here i am unsure whether the +c is multiplied by -1/a^2] assuming it is not gives:

-1/(u+1)= -y^2/(2*a^2)+c

substituting back for u gives

-1/((y/x)+1))=-y^2/(2*a^2)+c

[not sure if this is next step is right]

-(x/y)-1=-y^2/(2*a^2)+c

-x-y=(-y^3)/(2*a^2)+c

x=(y^3)/(2*(a^2))-y+c

thanks for your time

any help would be much appretiated

dooogle
 

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