Look at differential equations in the form
y′+p(x)y=q(x)yn(∗)
where p(x) and q(x) are continuous functions on
the interval we’re working on and n is a real number.
divide the differential equation by yn to get
y−ny′+p(x)y1−n=q(x)
We are now going to use the substitution
v=y1−n
to convert this into a differential equation in terms of v .
As we’ll see this will lead to a differential equation that we can solve. So
v′=(1−n)y−ny′
Now, plugging this as well as our substitution into the differential equation gives,
1−n1v′+p(x)v=q(x)
This is a linear differential equation that we can solve for v
and once we have this in hand we can also get the solution to the
original differential equation by plugging v
back into our substitution and solving for y .
So the equation is reduced to a linear equation.
If n=0 equation (*) is a linear equation, in our case
n=2 equation is not linear equation, but reduced to a linear equation.
Look at an example.
y′+x4y=x3y2
So, the first thing that we need to do is get this into
the “proper” form and that means dividing everything by y2 . Doing this gives
y−2y′+x4y−1=x3
The substitution and derivative that we’ll need here is
v=y−1v′=−y−2y′
With this substitution the differential equation becomes
−v′+x4v=x3
Here’s the solution to this differential equation
v′−x4v=−x3⇒μ(x)=e∫−x4dx=e−4ln∣x∣=x−4
∫(x−4v)′dx=∫−x−1dxx−4v=−ln∣x∣+c⇒v(x)=cx4−x4lnx
So, to get the solution in terms of y all we need to
do is plug the substitution back in. Doing this gives,
y−1=x4(c−lnx)
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