How do you do (a) and (b)? Bernoulli’s equation is an equation of the form y ′ = a(t)y + f(t)y n , where n can't be 0 or 1. (a) Using the substitution z = y 1−n , show that we can transform Bernoulli’s - QAmmunity.org

How do you do (a) and (b)? Bernoulli’s equation is an equation of the form y ′ = a(t)y + f(t)y n , where n can't be 0 or 1. (a) Using the substitution z = y 1−n , show that we can transform Bernoulli’s

asked Jul 27, 2020 142k views

1 Answer

Answer:

See solution below

Explanation:

(a) If n=0 or 1, the equation

(1)
y' = a(t)y + f(t)y^n

would be a simple linear differential equation. So, we can assume that n is different to 0 or 1.

Let's use the following substitution:

(2)
z=y^(n-1)

Taking the derivative implicitly and using the chain rule:

(3)
z'=(1-n)y^(-n)y'

Multiplying equation (1) on both sides by

(1-n)y^(-n)

we obtain the equation

(1-n)y^(-n)y' = (1-n)y^(-n)a(t)y+(1-n)y^(-n)f(t)y^n

reordering:

(1-n)y^(-n)y' = (1-n)y^(-n)ya(t)+(1-n)y^(-n)y^nf(t)

(1-n)y^(-n)y' = (1-n)y^(1-n)a(t)+(1-n)y^(0)f(t)

(1-n)y^(-n)y' = (1-n)y^(1-n)a(t)+(1-n)f(t)

Now, using (2) and (3) we get:

z'= (1-n)za(t)+(1-n)f(t)

which is an ordinary linear differential equation with unknown function z(t).

(b)

The equation we want to solve is

(4)
xy'+ y = x^4 y^3

Here, our independent variable is x (instead of t)

Assuming x different to 0, we divide both sides by x to obtain:

y'+(1)/(x)y = x^3 y^3

y' = -(1)/(x)y+x^3 y^3

Which is an equation of the form (1) with

a(x)=-(1)/(x)

f(x)=x^3

n=3

So, if we substitute

z=y^(-2)

we transform equation (4) in the lineal equation

(5)
z'=(2)/(x)z-2x^3

and this is an ordinary lineal differential equation of first order whose

integrating factor is

$e^{\int (-(2)/(x))dx}$

but

$e^{\int (-(2)/(x))dx}=e^{-2\int (dx)/(x)}=e^(-2ln(x))=e^{ln(x^(-2))}=x^(-2)=(1)/(x^2)$

Similarly,

$e^{\int ((2)/(x))dx}=x^2$

and the general solution of (5) is then

$z(x)=x^2\int ((-2x^3)/(x^2))dx+Cx^2=-2x^2\int xdx+Cx^2=\\\ -2x^2(x^2)/(2)+Cx^2=-x^4+Cx^2$

where C is any real constant

Reversing the substitution

z=y^(-2)

we obtain the general solution of (4)

$y=\sqrt{(1)/(z)}=\sqrt{(1)/(-x^4+Cx^2)}$

Attached there is a sketch of several particular solutions corresponding to C=1,4,6

It is worth noticing that the solutions are not defined on x=0 and for C<0

How do you do (a) and (b)? Bernoulli’s equation is an equation of the form y ′ = a-example-1

Yojimbo answered Jul 31, 2020

6.9k points

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