xy' = y − xe^(y/x)
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You should use the following substitution such that:
`y = x*f(x)`
Differentiating both sides with respect to x yields:
`(dy)/(dx) = (d(x*f(x)))/(dx)`
You need to use product rule to the right side such that:
`y' = (dy)/(dx) = f(x) + (x*d(f(x)))/(dx)`
You need to substitute `f(x) + (x*d(f(x)))/(dx)` for `y` ' and `x*f(x)` for `y` , in the given equation, such that:
`x(f(x) + (x*d(f(x)))/(dx)) =x*f(x)- xe^((x*f(x))/x)`
You need to factor out x to the right side such that:
`x(f(x) + (x*d(f(x)))/(dx)) = x*(f(x)- e^((x*f(x))/x))`
Equating like parts both sides such that:
`(x*d(f(x)))/(dx)) = -e^((f(x))/x)`
Dividing both sides by x yields:
`(d(f(x)))/(dx)) = (-e^((f(x))/x))/x`
You should divide both sides by `e^((f(x)))` such that:
`((d(f(x)))/(dx))/(e^((f(x)))) = -1/x`
Integrating both sides yields:
`int ((d(f(x)))/(dx))/(e^((f(x)))) dx= - int 1/x dx`
`- e^((-f(x)) = - ln|x| + c => e^(-f(x)) = ln|x| - c`
Taking logarithms both sides yields:
`ln(e^((-f(x)))) = ln(ln|x| - c)`
Using logarithmic identities yields:
`-f(x)ln e = ln(ln|x| - c) => -f(x) = ln(ln|x| - c)`
You may evaluate the solution to the given equation such that:
`y = x*f(x) => y = -x*(ln(ln|x| - c))`
Hence, evaluating the solution to the given first order nonlinear differential equation yields `y = -x*(ln(ln|x| - c)).`
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The y/x within the exponential looks problematic, so we can try substituting u = y/x
u' = y'/x - y/x^2
=> u'x^2 = xy' - y
Hence substituting into the original equation we get
u'x^2 = -xe^u => u'x = -e^u
We can arrange this as e^-u du = -1/x dx
so integrating each side gives
e^-u = log(x) + C
Substituing u = y/x back into the equation gives
e^-(y/x) = log(x) + C
-y/x = log(log(x) + C)
y = -x*log (log(x) + C)
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