Q.The order and degree of the differential equation $\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^{1/4}+x^{1/5}=0$, respectively, are:
(A) 2 and not defined
(B) 2 and 2
(C) 2 and 3
(D) 3 and 3

The key idea is simple: the order of a differential equation tells you how many arbitrary constants appear in its general solution. That’s a fundamental theorem — every time you integrate, you introduce one constant, and the order is the highest number of derivatives you need to undo.

So the real work is just finding the order of this equation. But the equation looks messy — there’s a fractional power. Let’s clean it up.

1. Rewrite the equation in standard form. The given equation is

To find the order, we need the highest derivative that appears after removing any radicals or fractional exponents. Raise both sides to the power $2/3$ to get rid of the $3/2$ exponent:

Now the left side has $\frac{d^4y}{dx^4}$ — that’s a fourth derivative. The right side has $\frac{d^3y}{dx^3}$ inside a fractional power, but that doesn’t lower the order; the highest derivative present is still the fourth.

2. Identify the order.

   The order of a differential equation is the highest order derivative that appears in the equation after it has been rationalized (no radicals or fractional powers on the derivative terms themselves). Here, $\frac{d^4y}{dx^4}$ is clearly the highest, so the order is 4.

3. Connect order to number of constants. …

An equation with two arbitrary constants yields a differential equation whose order equals the number of constants, so $k=2$.

For $k=2$ the curve is $x^2+y^2=c^2$ (one arbitrary constant $c$). Differentiate with respect to $x$: …

The given equation is $y = ax^2 - 2abx + ab^2$. Notice that the right-hand side is a perfect square in $x$:

So the family is simply $y = a(x-b)^2$ — a set of parabolas opening upward or downward (depending on $a$) with vertex at $(b,0)$. Two arbitrary constants means we need a second-order differential equation.

1. Differentiate once. From $y = a(x-b)^2$,

2. Differentiate again.

3. Eliminate $a$ and $b$. From the second derivative we have $a = \frac{y''}{2}$. Substitute into the first derivative:

So $x-b = \frac{y'}{y''}$.

4. Now use the original equation.

   $y = a(x-b)^2 = \frac{y''}{2} \left( \frac{y'}{y''} \right)^2 = \frac{(y')^2}{2y''}$.

   Multiply through: $2y y'' = (y')^2$. …

Conjugate-root theorem (rational/real coefficients).

• A polynomial with rational (hence real) coefficients that has a complex root $1+2i$ must also have its conjugate $1-2i$.
• With rational coefficients, an irrational root of the form $2-\sqrt{3}$ must also have its radical conjugate $2+\sqrt{3}$.
• $5$ is rational, so it needs no partner. …