Question Details
A bernoulli differential equation is one of the form dy/dx +P(x)y=Q(x)y^n (*)
Observe that if n=0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u=y^1-n transforms the Bernoulli equation into the linear equation du/dx + (1-n)P(x)u=(1-n)Q(x)
Solve the equation using an appropriate substitution to solve the equation
xy’ +y=-3xy^2 and y(1)=2
I can't seem to solve (B) and (D)
a) The differential ewuation can be written in the form (*) with P(x)____?, Q(x)_____? And n=?
Answer: P(x)=1/x , Q(x)=-3, n=2
b) The substitution u=__y^-1_____ will transform it into the linear equation
du/dx+ _?____u= ?
Answer:du/dx+ _?____u= -3
c) Using the substitution in part(b), we rewrite the initial condition in terms of X and u: u(1)= ?
Answer:u(1)=0.5
d) Solve the linear equation in part (b) and find the solution that satisfies the initial condition in part(c).
U(x)=???
Answers & Comments
Verified answer
y' + (1/x) y = -3y², y(1) = 2.
Putting u = y^(1 - 2) gives
u = y^(-1) ==> du/dx = -y^(-2) dy/dx.
We get a couple of things from this change of variables:
dy/dx = -y² du/dx, and from u = y^(1 - 2) = y y^(-2), y = y²u.
Substituting all of this into the original equation produces
-y² du/dx + (1/x) y² u = -3y².
Cancelling -y² from both sides yields a linear equation in u.
du/dx - 1/x u = 3.
The integrating factor is µ = exp(∫ -1/x dx) = 1/x.
d/dx [u/x] = 3/x ==> u/x = 3ln|x| + C ==> u(x) = 3x ln(x) + Cx.
From y(1) = 2, u(1) = 1/y(1) = 1/2.
u(1) = 3ln1 + C = 1/2 ==> C = 1/2.
So u(x) = 3x ln(x) + x/2.
Take the reciprocal to get y.
y = 2/(6xln(x) + x).
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