# Solutions to Various First Order ODEs

Solve using integrating factor and Bernoullis.

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#### Solution Preview

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, ...

#### Solution Summary

We solve various types of first order ordinary differential equations.