Differential Equation Probability

The order of the differential equation : $\large{\color{#3D99F6}{x^{3}\frac{d^{3}y}{dx^{3}}+x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y=0}}$ is equal to $3$ since order of a differential equation is equal to the order of highest order derivative in it.

The degree of a differential equation is defined only when it is expressed as a polynomial in derivatives . $\large{\color{#3D99F6}{{ y }^{ 2 }=2\frac { dy }{ dx } \left( x+\sqrt [ 5 ]{ \frac { dy }{ dx } } \right)}}$

This is not a polynomial in derivatives as we have a $5th$ root of derivative involved here. So, we make it a polynomial in derivatives as follows:

$\large{\color{#3D99F6}{\frac{y^{2}}{2y'}-x=\left (\frac{dy}{dx} \right )^{1/5}}}$

Now, taking 5th power on both sides:

$\large{\color{#3D99F6}{\left (\frac{y^{2}}{2y'}-x \right )^{5}= y'}}$

$\large{\color{#3D99F6}{\left (\frac{y^{2}-2xy'}{2y'} \right )^{5}= y'}}$

$\large{\color{#3D99F6}{\frac{(y^{2}-2xy')^{5}}{32(y')^{5}} = y'}}$

$\large{\color{#3D99F6}{(y^{2}-2xy')^{5}=32(y')^{6}}}$

Clearly, maximum power of $y'$ will be $5$ on LHS and it is $6$ on RHS. Hence, the degree of this differential equation is $6$ .

So, we need to find the probability that out of $3$ numbers chosen from the given set, the minimum no. obtained is $3$ provided that the maximum no. obtained is $6$ .

Let A denote the event that min. no. obtained is $3$ and B denote the event that max. no. obtained is $6$ . So, $\large{P\left ( \frac{A}{B} \right )}$ is what we need to find. Using the conditional probability:

$\large{P\left ( \frac{A}{B} \right )=\frac{P(A\cap B)}{P(B)}}$

Let us find the favourable cases to the events $(A\cap B)$ and $B$ .

$(A\cap B)$ denotes that the min no obtained is 3 and max no obtained is 6. So, out of the 3 numbers, 2 numbers must be 3 and 6 and the 3rd number can be either 4 or 5. Hence, $n(A \cap B)=2$

Now, $n(B)=^{5}C_{2}=10$ because B denotes that max no obtained is 6, hence out of 3 no., one must be 6 and the rest 2 numbers can takes values from ${1,2,3,4,5}$ i.e. we have to choose 2 values from 5.

Let $n(S)$ denote the no. of elements in the sample space.

Hence, $\large{P\left ( \frac{A}{B} \right )=\frac{n(A \cap B)/n(S)}{n(B)/n(S)}=\frac{2}{10}=0.2}$