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# Solutions to Various First Order ODEs

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Solve using integrating factor and Bernoullis.

https://brainmass.com/math/integrals/solutions-various-first-order-odes-430879

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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, exponentiating both sides we obtain

mu(y) = 1/y^5.

Plugging this integrating factor into (1), we obtain

(2) 1/y^4 dx - 4(x/y^5 + y) dy = 0.

Thus we have

M(x,y) = 1/y^4
N(x,y) = -4x/y^5 - 4y

We know (2) is exact, so its solution is given by

C = f(x,y) = integral(M(x,y) dx)
= x/y^4 + g(y)

for some function g(y). We have

f_y = N(x,y)
-4x/y^5 + g'(y) = -4x/y^5 - 4y

whence

g'(y) = -4y
g(y) = -2y^2

Therefore the solution to the differential equation is

x/y^4 - 2y^2 = C.

4. We have

L di/dt + Ri = E

with initial condition i(0) = i_0.

Dividing both sides by R, we obtain

di/dt + R/L i = E/L.

This is a linear ODE with integrating factor given by

mu(t) = exp(integral(R/L dt))
= e^(Rt/L).

Multiplying both sides by mu(t), we obtain

e^(Rt/L) di/dt + R/L i(t) e^(Rt/L) = E/L e^(Rt/L)
d/dt(i(t) e^(Rt/L)) = E/L e^(Rt/L).

Integrating both sides, we obtain

i(t) e^(Rt/L) = E/R e^(Rt/L) + C.

Finally, dividing both sides by e^(Rt/L), we obtain the general solution

i(t) = E/R + C e^(-Rt/L).

Plugging in the initial condition we obtain

i(0) = E/R + C = i_0

whence

C = i_0 - E/R.

Therefore the solution to the differential equation with the initial condition is

i(t) = E/R + (i_0 - E/R) e^(-Rt/L).

5. We have

(x + 1) dy/dx + y = ln x

with initial condition

y(1) = 10.

Dividing both sides by x + 1, we obtain

dy/dx + y/(x+1) = (ln x)/(x+1).

This is a linear ODE with integrating factor given by

mu(x) = exp(integral(dx/(x+1)))
= exp(ln(x+1))
= x + 1.

Multiplying both sides by the integrating factor, we obtain

(x + 1) dy/dx + y = ln x
((x + 1) y)' = ln x.

Integrating both sides we obtain

(x + 1) y = x ln x - x + C

whence

y(x) = (x ln x - x + C)/(x + 1).

Plugging in the initial condition we obtain

10 = y(1) = (C - 1)/2

whence

20 = C - 1
21 = C.

Therefore the solution satisfying the initial condition is

y(x) = (x ln x - x + 21)/(x + 1).

6. We have

dy/dx = y(xy^3 - 1)

which we may write as

dy/dx + y = xy^4.

This is a Bernoulli ODE with n = 4. To solve it, we first make the change of variables

w = 1/y^3

whence

y = w^(-1/3)
y' = -1/3 w^(-4/3) w'

so our differential equation becomes

-1/3 w^(-4/3) w' + w^(-1/3) = xw^(-4/3).

Multiplying both sides by -3w^(4/3), we obtain the linear ODE

w' - 3w = -3x.

The integrating factor for this ODE is given by

mu(x) = exp(integral(-3 dx)) = e^(-3x).

Multiplying both sies by mu(x), we obtain

(w' - 3w) e^(-3x) = -3x e^(-3x)
(w e^(-3x))' = -3x e^(-3x).

Integrating both sides, we obtain

w e^(-3x) = (x + 1/3)e^(-3x) + C.

whence

w = 1/y^3 = x + 1/3 + C e^(3x).

so

y = (x + 1/3 + C e^(3x))^(-1/3).

7. We have

x^2 dy/dx - 2xy = 3y^4

with initial condition y(1) = 1/2.

Dividing both sides by x^2 we obtain

dy/dx - 2y/x = 3y^2.

This is a Bernoulli ODE with n = 2, so we make the substitution

w = 1/y

whence

y = 1/w
y' = -w'/w^2

and our ODE becomes

-w'/w^2 - 2/(wx) = 3/w^2.

Multiplying both sides by -w^2 we obtain

w' + 2w/x = -3.

This is a linear ODE with integrating factor

mu(x) = exp(integral(2 dx/x)) = e^(2 ln x) = x^2.

Thus we have

(wx^2)' = -3 x^2
wx^2 = -x^3 + C
w = 1/y = -x + C/x^2
y = 1/(C/x^2 - x)
= x^2/(C - x^3).

Plugging in the initial condition we have

1/2 = y(1) = 1/(C - 1)

whence C = 3. Thus the solution to the ODE satisfying the initial condition is

y(x) = x^2/(3 - x^3).

8. We have

xy(1 + xy^2) dy/dx = 1

with initial condition y(1) = 0.

We make the substitution

u = xy^2

whence

du/dx = y^2 + 2xy dy/dx

so

xy dy/dx = 1/2(du/dx - y^2)
= 1/2(du/dx - u/x).

Thus we have

1/2(1 + u)(du/dx - u/x) = 1,

whence

du/dx - u/x = 2(1 + u)
du/dx - u(2 + 1/x) = 2.

This is a linear ODE with integrating factor

mu(x) = exp(integral((-2 - 1/x)dx))
= exp(-2x - ln x)
= e^(-2x)/x.

Thus we have

(ue^(-2x)/x)' = 2e^(-2x)/x
ue^(-2x)/x = 2 integral(e^(-2x)/x dx) = 2 Ei(2x) + C,

where Ei is the exponential integral function. Thus we have

u(x) = xy^2(x) = (2 Ei(2x) + C)x e^(2x)

whence

y(x) = e^x sqrt(2 Ei(2x) + C).

Plugging in the initial condition, we have

0 = y(1) = e sqrt(2 Ei(1) + C),

whence C = -2 Ei(1) and the solution to the ODE satisfying the initial condition is

y(x) = e^x sqrt(2(Ei(2x) - Ei(1)).

References:

http://en.wikipedia.org/wiki/Bernoulli_differential_equation
http://en.wikipedia.org/wiki/Exponential_integral

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