Simplifying Variables / Exponents.?

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x∙y' - y = √(x² + y²)
<=>
y' = √(x² + y²)/x + y/x
<=>
y' = √( (x² + y²)/x² ) + y/x
<=>
y' = √(1 + (y/x)²) + y/x

So you have a ordinary first order differential equation of the form
y' = F(y/x)
a so called first order homogeneous differential equation. You can convert is to a separable first order equation by substituting y/x.
Let
u = y/x
<=>
y = x∙u
=>
y' = x∙u' + u

y' = √(1 + (y/x)²) + y/x
<=>
x∙u' + u = √(1 + u²) + u
<=>
u' = du/dx = (1/x)∙√(1 + u²)
<=>
(1/√(1 + u²)) du = (1/x) dx
=>
∫ (1/√(1 + u²)) du = ∫ (1/x) dx

Left hand side integral can be solved by trigonometric substitution:
u = tan(t)
=>
du = (tan²(t) + 1) dt
=>
∫ (1/√(1 + u²)) du
= ∫ (tan²(t) + 1)/√(1 + tan²(t)) dt
= ∫ √(tan²(t) + 1) dt
< since tan²(t) + 1 = sec²(t)
= ∫ sec(t) dt
= ln(sec(t) + tan(t))
= ln(√(tan²(t) + 1) + tan(t))
= ln(√(u² + 1) + u)

Hence,
∫ (1/√(1 + u²)) du = ∫ (1/x) dx
<=>
ln(√(u² + 1) + u) = ln(x) + c
<=>
√(u² + 1) + u = e^( ln(x) + c ) = e^(c)∙e^(ln(x)) = e^(c)∙x
set C = e^(c)
√(u² + 1) + u = C∙x
<=>
√((y/x)² + 1) + (y/x) = C∙x
<=>
√(y² + x²) + y = C∙x²
<=>
√(y² + x²) = C∙x² - y
<=>
y² + x² = (C∙x² - y)²
<=>
y² + x² = (C²∙x⁴ - 2∙C∙x²∙y + y²
<=>
2∙C∙x²∙y = (C²∙x⁴ - x²)
<=>
y = (C/2)∙x² - 1/(/2∙C)
set C' = 2∙C
y = 4∙C'∙x² - 1/C'...Show more