Hello everyone !! I have this problem: "DEMOSTRATE that the differential equation $xdy-ydx=tan^{-1}(y/x)dx$ can be solved by using the substitution of $y=vx$ So i proceed to solve it:
y=vx
Then
dy = vdx + xdv
And so, for the original differential equation:
\(\displaystyle x(vdx+xdv) - xvdx = tan^{-1}(\frac{vx}{x})dx\)
Simplifiyng and re-ordening i get:
\(\displaystyle x^2dv-tan^{-1}(v)dx=0\) (Separable differential equation)
Solving:
\(\displaystyle \int \dfrac{1}{tan^{-1}(v)}dv=\int \dfrac{1}{x^2}dx\)
\(\displaystyle \int \dfrac{1}{tan^{-1}(v)}dv=-\dfrac{1}{x}+C\)
But to complete the problem i need return to the original variable $y$ but i dont know how to solve the left-side integral.... perhaps i'm doing something wrong... can someone help me please???
y=vx
Then
dy = vdx + xdv
And so, for the original differential equation:
\(\displaystyle x(vdx+xdv) - xvdx = tan^{-1}(\frac{vx}{x})dx\)
Simplifiyng and re-ordening i get:
\(\displaystyle x^2dv-tan^{-1}(v)dx=0\) (Separable differential equation)
Solving:
\(\displaystyle \int \dfrac{1}{tan^{-1}(v)}dv=\int \dfrac{1}{x^2}dx\)
\(\displaystyle \int \dfrac{1}{tan^{-1}(v)}dv=-\dfrac{1}{x}+C\)
But to complete the problem i need return to the original variable $y$ but i dont know how to solve the left-side integral.... perhaps i'm doing something wrong... can someone help me please???
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