The Bernoulli Differential Equation

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:

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 = e^{x6 + C}

u = e^{(x6 + c)} + 1

Substitute back y = u⁽⁻¹⁶⁾

y = ( e^{(x6 + c)} + 1 )⁽⁻¹⁶⁾

Solved!

And we get these example curves:

Example 2: Solve

dydx − yx = y⁹

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

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!

And we get this nice family of curves:

Example 3: Solve

dydx + 2yx = x²y²sin(x)

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

vdwdx + wdvdx − 2vwx = −x²sin(x)

Step 4: Factor the parts involving w.

vdwdx + w(dvdx − 2vx) = −x²sin(x)

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

dvdx − 2vx = 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)

Which looks like this (example values of C):