Solutions to Several First Order ODEs
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Solve 1st order DE, please include explaination of process, if possible.
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Solution Summary
We use various methods to solve several first order ordinary differential equations.
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1. We have
dy/dx = (x + 3y)/(3x + y) = (1 + 3y/x)/(3 + y/x).
Let u = y/x, so y = ux and dy/dx = u + x du/dx. Then we have
u + x du/dx = (1 + 3u)/(3 + u)
so
x du/dx = (1 + 3u)/(3 + u) - u
= (1 - u^2)/(3 + u).
Thus we have
(3 + u)/(1 - u^2) du = dx/x.
Integrating both sides we find
ln((1 + u)/(1 - u)^2) = ln x + C1.
Exponentiating both sides we have
(1 + u)/(1 - u)^2 = Cx.
Substituting y back into the equation and simplifying, we find
x + y = C(x - y)^2.
2. We have
-y dx + (x + sqrt(xy)) dy = 0,
whence
dy/dx = y/(x + sqrt(xy)) = u/(1 + sqrt(u))
where u = y/x. Thus we have
u + x du/dx = u/(1 + sqrt(u))
whence
x du/dx = u/(1 + sqrt(u)) - u
= u(1 - 1 - sqrt(t))/(1 + sqrt(u))
= -u sqrt(u)/(1 + sqrt(u)).
Thus we have
(1 + u^(1/2)) / u^(3/2) du = dx/x
[u^(-3/2) + u^-1] du = dx/x.
Integrating both sides we find
-2 u^(-1/2) + ln u = ln x + C.
Substituting y back into this equation, we have
-2 sqrt(x/y) + ln y - ln x = ln x + C
whence
-2 sqrt(x/y) + ln y - 2 ln x = C.
3. We have
dy/dx = y/x + x/y
which is clearly homogeneous. Substituting u = y/x, we find
u + x du/dx = u + 1/u
whence
x du/dx = 1/u
from which it follows that
u du = dx/x.
Integrating both sides, we find
1/2 u^2 = ln x + C.
Substituting y back in, we find
y^2/(2x^2) = ln x + C
whence
y^2 = 2x^2 ln x + C
so the solution to the DE is
y = +/- sqrt(2x^2 ln x + C).
4. We have
(y + x cot(y/x)) dx - x dy = 0,
from which it follows that
(y/x + (x/y) cot(x/y)) dx - dy = 0,
which is clearly homogeneous. From the substitution u = y/x, we have
(u + (cot u)/u) dx - (u dx + x du) = 0,
from which we obtain
(cot u)/u dx = x du,
from which ...
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