Answer to Question #105497 in Differential Equations for khushi

Question

State whether the following statements are true or false. Justify yourself with the help of
a short proof or a counter example.
i) y'+ P(x) y = Q(x) y^2 is a linear equation for integral values of n.

Answer 1

Look at differential equations in the form

$y' + p\left( x \right)y = q\left( x \right){y^n}\quad (*)$

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 $y^n$  to get

${y^{ - n}}\,y' + p\left( x \right){y^{1 - n}} = q\left( x \right)$

We are now going to use the substitution

$v = {y^{1 - n}}$

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

$v' = \left( {1 - n} \right){y^{ - n}}y'$

Now, plugging this as well as our substitution into the differential equation gives,

$\frac{1}{{1 - n}}v' + p\left( x \right)v = q\left( x \right)$

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.

$y' + \frac{4}{x}y = {x^3}{y^2}$

So, the first thing that we need to do is get this into

the “proper” form and that means dividing everything by $y^2$ . Doing this gives

${y^{ - 2}}\,y' + \frac{4}{x}{y^{ - 1}} = {x^3}$

The substitution and derivative that we’ll need here is

$v = {y^{ - 1}}\hspace{0.25in}v' = - {y^{ - 2}}y'$

With this substitution the differential equation becomes

$- v' + \frac{4}{x}v = {x^3}$

Here’s the solution to this differential equation

$v' - \frac{4}{x}v = - {x^3}\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\\\mu \left( x \right) = {{\bf{e}}^{\int{{ - \,\,\frac{4}{x}\,dx}}}} = {{\bf{e}}^{ - 4\,\,\ln \left| x \right|}} = {x^{ - 4}}$

$\int (x^{ - 4}v)' dx = \int{{ - {x^{ - 1}}\,dx}}\\ {x^{ - 4}}v = - \ln \left| x \right| + c\hspace{0.25in} \Rightarrow \\ \hspace{0.25in}v\left( x \right) = c{x^4} - {x^4}\ln x$

So, to get the solution in terms of $y$  all we need to

do is plug the substitution back in. Doing this gives,

${y^{ - 1}} = {x^4}\left( {c - \ln x} \right)$