Question

Is y+P(x)y=Q(x)yⁿ a linear equation for integral value of n. Justify your answer.

Answer 1

SOLUTION:-

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

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

$y ^{−n} y ′ +p(x)y^{1−n} =q(x)$

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

$v^ ′ =(1−n)y ^ {−n}$

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

$\frac{1}{1−n} ​ v ^ ′ +p(x)v=q(x)$

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.