# Stuck trying to integrate a differential equation

## Homework Statement

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

where a is a constant

express answer in the form x=f(y)

## 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

dooogle

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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

$$\frac{du}{dx}=1+ \frac{a^2}{u^2}$$

$$\Leftrightarrow \frac{du}{dx}=\frac{u^2 + a^2}{u^2}$$

$$\Rightarrow \int \frac{u^2 du}{u^2 + a^2} = \int dx$$

This integrand doesn't seem to agree with yours.

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

Char. Limit
Gold Member
Try using the opposite order. Differentiate u2 and integrate 1/(u2+a2).

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

Char. Limit
Gold Member
You integrated by parts incorrectly. You should get 2u/(u2+a2) as the second integral, not what you have.

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