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Solutions to Several First Order ODEs

Solve 1st order DE, please include explaination of process, if possible.

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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 ...

Solution Summary

We use various methods to solve several first order ordinary differential equations.

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