Question #105497
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.
1
Expert's answer
2020-03-24T09:44:59-0400

Look at differential equations in the form

y+p(x)y=q(x)yn()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 yny^n  to get

yny+p(x)y1n=q(x){y^{ - n}}\,y' + p\left( x \right){y^{1 - n}} = q\left( x \right)

We are now going to use the substitution

v=y1nv = {y^{1 - n}}

to convert this into a differential equation in terms of vv .

As we’ll see this will lead to a differential equation that we can solve. So

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

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

11nv+p(x)v=q(x)\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 vv

and once we have this in hand we can also get the solution to the

original differential equation by plugging vv

 back into our substitution and solving for yy .

So the equation is reduced to a linear equation.

If n=0n=0 equation (*) is a linear equation, in our case

n=2n=2 equation is not linear equation, but reduced to a linear equation.

Look at an example.

y+4xy=x3y2y' + \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 y2y^2 . Doing this gives

y2y+4xy1=x3{y^{ - 2}}\,y' + \frac{4}{x}{y^{ - 1}} = {x^3}

The substitution and derivative that we’ll need here is

v=y1v=y2yv = {y^{ - 1}}\hspace{0.25in}v' = - {y^{ - 2}}y'

With this substitution the differential equation becomes

v+4xv=x3- v' + \frac{4}{x}v = {x^3}

Here’s the solution to this differential equation

v4xv=x3μ(x)=e4xdx=e4lnx=x4v' - \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}}

(x4v)dx=x1dxx4v=lnx+cv(x)=cx4x4lnx\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 yy  all we need to

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

y1=x4(clnx){y^{ - 1}} = {x^4}\left( {c - \ln x} \right)


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