The Bernoulli Differential Equation

How to solve this special first order differential equation

A Bernoulli equation has this form:

dydx + P(x)y = Q(x)yn
where n is any Real Number but not 0 or 1

When n = 0 the equation can be solved as a First Order Linear Differential Equation.

When n = 1 the equation can be solved using Separation of Variables.

For other values of n we can solve it by substituting

u = y1−n

and turning it into a linear differential equation (and then solve that).


Example 1: Solve

dydx + x5 y = x5 y7

It is a Bernoulli equation with P(x)=x5, Q(x)=x5, and n=7, let's try the substitution:

u = y1−n

u = y−6

In terms of y that is:

y = u(−16)

Differentiate y with respect to x:

dydx = −16 u(−76) dudx

Substitute dydx and y into the original equation  dydx + x5 y = x5 y7

−16u(−76) dudx + x5u(−16) = x5u(−76)

Multiply all terms by −6u(76)

dudx − 6x5u = −6x5

The substitution worked! We now have an equation we can hopefully solve.

Simplify:

dudx = 6x5u − 6x5

dudx = (u−1)6x5

Using separation of variables:

duu−1 = 6x5 dx

Integrate both sides:

1u−1 du  = 6x5 dx

Gets us:

ln(u−1) = x6 + C

u−1 = ex6 + C

u = e(x6 + c) + 1

Substitute back y = u(−16)

y = ( e(x6 + c) + 1 )(−16)

Solved!

Let's look again at that substitution we did above. We started with:

dydx + x5y = x5y7

And ended with:

dudx − 6x5u = −6x5

In fact, in general, we can go straight from

dydx + P(x)y = Q(x)yn
n is not 0 or 1

to:

dudx + (1−n)uP(x) = (1−n)Q(x)

Then solve that and finish by putting back y = u(−1n−1)

Let's do that in the next example.

Example 2: Solve

dydxyx = y9

It is a Bernoulli equation with n = 9, P(x) = −1x and Q(x) = 1

Knowing it is a Bernoulli equation we can jump straight to this:

dudx + (1−n)uP(x) = (1−n)Q(x)

Which, after substituting n, P(X) and Q(X) becomes:

dudx + 8ux = −8

Now let's try to solve that.

Unfortunately we cannot separate the variables, but the equation is linear and is of the form dudx + R(X)u = S(x) with R(X) = 8x and S(X) = −8

Which we can solve with steps 1 to 9:

Step 1: Let u=vw

Step 2: Differentiate u = vw

dudx = vdwdx + wdvdx

Step 3: Substitute u = vw and dudx = v dwdx + w dvdx into dudx + 8ux = −8:

vdwdx + wdvdx + 8vwx = −8

Step 4: Factor the parts involving w.

vdwdx + w(dvdx + 8vx) = −8

Step 5: Set the part inside () equal to zero, and separate the variables.

dvdx + 8vx = 0

dvv = −8dxx

Step 6: Solve this separable differential equation to find v.

dvv = − 8dxx

ln(v) = ln(k) − 8ln(x)

v = kx−8

Step 7: Substitute v back into the equation obtained at step 4.

kx−8 dwdx = −8

Step 8: Solve this to find v

kx−8 dw = −8 dx

k dw = −8x8 dx

k dw = −8x8 dx

kw = −89x9 + C

w = 1k( −89 x9 + C )

Step 9: Substitute into u = vw to find the solution to the original equation.

u = vw = kx−8k( −89 x9 + C )

u = x−8 ( − 89 x9 + C )

u = −89x + Cx−8

Now, the substitution we used was:

u = y1−n = y−8

Which in our case means we need to substitute back y = u(−18) :

y = ( −89 x + c x−8 ) (−18)

Done!

Example 3: Solve

dydx + 2yx = x2y2sin(x)

It is a Bernoulli equation with n = 2, P(x) = 2x and Q(x) = x2sin(x)

We can jump straight to this:

dudx + (1−n)uP(x) = (1−n)Q(x)

Which, after substituting n, P(X) and Q(X) becomes:

dudx2ux = − x2sin(x)


In this case, we cannot separate the variables, but the equation is linear and of the form dudx + R(X)u = S(x) with R(X) = −2x and S(X) = −x2sin(x)

Solve the steps 1 to 9:

Step 1: Let u=vw

Step 2: Differentiate u = vw

dudx = vdwdx + wdvdx

Step 3: Substitute u = vw and dudx = vdwdx + wdvdx into dudx2ux = −x2sin(x)

vdwdx + wdvdx2vwx = −x2sin(x)

Step 4: Factor the parts involving w.

vdwdx + w(dvdx2vx) = −x2sin(x)

Step 5: Set the part inside () equal to zero, and separate the variables.

dvdx2vx = 0

1vdv = 2xdx

Step 6: Solve this separable differential equation to find v.

1v dv = 2x dx

ln(v) = 2ln(x) + ln(k)

v = kx2

Step 7: Substitute u back into the equation obtained at step 4.

kx2dwdx = −x2sin(x)

Step 8: Solve this to find v.

k dw = −sin(x) dx

k dw = −sin(x) dx

kw = cos(x) + C

w = cos(x) + Ck

Step 9: Substitute into u = vw to find the solution to the original equation.

u = kx2cos(x) + Ck

u = x2(cos(x)+C)

Finally we substitute back y = u-1

y = 1x2 (cos(x)+C)

Tada!

 

The Bernoulli Equation is attributed to Jacob Bernoulli (1655-1705), one of a family of famous Swiss mathematicians.


 
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