Answer to Question #107195 in Differential Equations for Nikesh gautam pandit ji

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^n is a linear equation for integral values of n.

ii) y = 0, is a singular solution of the differential equation 27y-8(dy/dx)^3=0

iii) Equation x^2 ( y − px) = yp^2 is reducible to clairaut’s form

Answer 1

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^n is a linear equation for integral values of n. 

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

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

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

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

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

$−v' + {4 \above{2pt} x} ​ v=x ^ 3$

Here’s the solution to this differential equation

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

$∫(x^{ −4} v) 'dx=−ln∣x∣+c⇒ v(x)=cx^ 4−x ^4lnx$

So, to get the solution in terms of y

 all we need to

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

$y ^{ −1} =x ^ 4 (c−\ln{x})$

ii) y = 0, is a singular solution of the differential equation 27y-8(dy/dx)^3=0 

General solution to the equation is: $y=(x+C)^\frac{3}{2}$ .

The solution y=0 cannot be obtained from the general solution for any value of constant with C,

hence y=0  is a singular solution of this equation.

iii) Equation x^2 ( y − px) = yp^2 is reducible to clairaut’s form

Let's put $X=x^2;\quad Y=y^2$ then $​$ $p={x\above{2pt} y}{dY \above{2pt} dX}$ .. Let's denote $P= {dY\above{2pt} dX}$

Then the equation can be rewritten in form

$X(y−x^ 2{P\above{2pt} y}=y{x^2\above{2pt} y^2}P^2$

By multiplying both sides with y, we assume

$X(y^ 2 −x ^ 2 P)=x^ 2 P ^ 2$

or

$X(Y−XP)=XP ^ 2$

Therefore

$Y=XP+P ^2$

which is now in Clairaut’s form

The solution got by just replacing P by constant c.

Hence

$Y=cX+c^2$

$or \, y^2=cx^2+c^2$