Solutions to Various First Order ODEs
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Solve using integrating factor and Bernoullis.
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Solution Summary
We solve various types of first order ordinary differential equations.
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1. We have
x^2 y' + xy = 1.
Dividing both sides by x^2 we obtain
y' + y/x = 1/x^2.
This is a linear ODE with integrating factor given by
mu(x) = exp(integral(dx/x)) = e^(ln x) = x.
Multiplying both sides by the integrating factor we obtain
xy' + y = 1/x
(xy)' = 1/x.
Integrating both sides we obtain
xy = ln x + C.
Finally, dividing both sides by x we obtain the solution
y = (ln x + C)/x.
2. We have
cos x dy/dx + y sin x = 1.
Dividing both sides by cos x, we obtain
dy/dx + y tan x = sec x.
Once again, this is a linear ODE. The integrating factor is given by
mu(x) = exp(integral(tan x dx))
= exp(ln(sec x))
= sec x.
Multiplying both sides by mu(x) we obtain
sec x dy/dx + y tan x sec x = sec^2 x
(y sec x)' = sec^2 x.
Integrating both sides we obtain
y sec x = tan x + C.
Finally, multiplying both sides by cos x, we obtain the solution
y = sin x + C cos x.
3. We have
y dx - 4(x + y^6) dy = 0.
We look for an integrating factor mu(y) which makes the above equation exact, i.e.
(1) y mu(y) dx - 4(x + y^6) mu(y) = 0
should be exact. We let
M(x,y) = y mu(y)
N(x,y) = -4(x + y^6) mu(y)
We have
M_y = y mu'(y) + mu(y)
N_x = -4 mu(y)
Now in order for (1) to be exact, we need M_y = N_x, i.e.
y mu'(y) + mu(y) = -4 mu(y)
y mu'(y) = -5 mu(y)
Dividing both sides by y mu(y), we obtain
mu'(y)/mu(y) = -5/y
(ln mu(y))' = -5/y
Integrating both sides we obtain
ln mu(y) = -5 ln y
Finally, ...
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