Q.Determine the order and degree, if defined, of the differential equation: $y' + 5y = 0$

Concept & Intuition

The degree of a differential equation is the power of the highest-order derivative after the equation has been made a polynomial in all its derivatives (all radicals and fractional powers removed).

Step-by-step solution

Start from

1. Square both sides to remove the square root:

2. Isolate the cube-root term and cube both sides to remove the $1/3$ power:

3. The equation is now polynomial in the derivatives. The highest-order derivative is $\dfrac{d^2y}{dx^2}$ (order $2$), and its highest power on the right side is $\left(\frac{d^2y}{dx^2}\right)^{2\times 3}=\left(\frac{d^2y}{dx^2}\right)^{6}$.

So the degree $=6$.

Given equation.

Express $\frac{dx}{dy}$ in terms of $\frac{dy}{dx}$. Since $\dfrac{dx}{dy}=\dfrac{1}{\,dy/dx\,}$, write $p=\dfrac{dy}{dx}$:

Make it a polynomial in the derivative (degree is defined only after removing radicals and fractional powers of the derivatives). Multiply through by $p$:

Read off order and degree.

• Order = highest derivative present = first derivative $\Rightarrow$ order $1$.
• Degree = power of that highest derivative in the polynomial form = $3$.

Step 1 — Identify the order

$2y' + (y'')^2 = \sqrt{y'' - 3}$. The highest-order derivative present is $y''$, so the order is $2$.

Step 2 — Clear the radical to find the degree

The degree of a differential equation is defined only after it is written as a polynomial in the derivatives, free of radicals/fractional powers. Since $y''$ sits inside a square root on the right, square both sides:

Step 3 — Read off the degree

Expanding the left side gives a term $\left((y'')^2\right)^2 = (y'')^4$, so the highest power of the highest-order derivative $y''$ is $4$. Hence the degree is $4$.

Step 4 — Add the squares

The key idea: Degree is defined only when the differential equation is a polynomial in the derivatives. Fractional exponents must be eliminated by raising both sides to a suitable power, but we must be careful not to introduce extraneous roots or change the equation’s essential nature.

Why this approach works

The given equation is:

Both sides have fractional exponents. To find the degree, we need the highest-order derivative (here $\frac{d^2y}{dx^2}$) to appear with an integer exponent, and the equation must be a polynomial in all derivatives. The trick: find the least common multiple of the denominators of the exponents (4 and 3) and raise both sides to that power.

Step-by-step solution

1. Identify the highest-order derivative

   The highest derivative present is $\frac{d^2y}{dx^2}$ (order 2). The equation also contains $\frac{dy}{dx}$ (order 1), but the degree is determined by the highest-order derivative.

2. Clear the fractional exponents

   The exponents are $\frac{3}{4}$ on the left and $\frac{1}{3}$ on the right. The LCM of 4 and 3 is 12. Raise both sides to the 12th power:

This simplifies to:

because $\frac{3}{4} \times 12 = 9$ and $\frac{1}{3} \times 12 = 4$.

3. Check polynomial form

   The left side is $\left(1 + (y')^2\right)^9$, which is a polynomial in $y'$ (since expanding gives integer powers). The right side is $(y'')^4$, a pure power. The equation is now a polynomial in the derivatives: no fractional exponents remain.

4. Read off the degree

   The highest-order derivative is $y''$, and its highest power in the polynomial equation is 4. Therefore, the degree of the differential equation is 4.

The key idea: Order is defined as the highest order derivative that appears in the equation after it has been cleared of radicals and fractions. We must first simplify the given expression to see what derivatives are actually present.

Let’s work through it step by step.

1. Rewrite the given equation The equation is

The expression inside the derivative is $\left(\frac{dy}{dx}\right)^3$, which is a function of the first derivative only.

2. Apply the chain rule to differentiate Let $p = \frac{dy}{dx}$. Then the equation becomes

Using the chain rule:

So the equation simplifies to

3. Identify the highest derivative

   In the simplified form, we see $\frac{d^2y}{dx^2}$ (the second derivative) explicitly. There is no third derivative or higher. Therefore, the highest order derivative present is 2.

4. Check for any hidden higher derivatives

   Could the original form hide a third derivative? No — the outermost operation is a single derivative of a function of the first derivative. Differentiating once can at most raise the order by one, so the maximum possible order here is 2. No further differentiation is implied.

Thus, the order of the differential equation is 2.

Start from $\log\left(\frac{dy}{dx}\right)^{1/2}=5x+4y$, i.e. $\frac{1}{2}\log\frac{dy}{dx}=5x+4y$.

Multiply by 2 and exponentiate:

The right side is a function of $x,y$ only (no derivatives). The equation is now polynomial in the highest-order derivative $\frac{dy}{dx}$, which occurs to the first power. Therefore the degree is $1$.

We are given the differential equation:

Concept and intuition:

The order of a differential equation is the highest derivative present. The degree is the power of that highest derivative after the equation has been made polynomial in all derivatives (i.e., no fractional powers, no derivatives in denominators). Here, the term $\frac{4(d^2y/dx^2)^3}{d^3y/dx^3}$ has a derivative in the denominator, so we must first clear that denominator to find the true degree.

1. Identify the highest-order derivative.

   The derivatives appearing are $\frac{d^2y}{dx^2}$ (order 2) and $\frac{d^3y}{dx^3}$ (order 3). The highest is order 3. So the order is $3$.

2. Rewrite to remove the denominator.

   Multiply the entire equation by $\frac{d^3y}{dx^3}$ to eliminate the fraction:

3. Bring all terms to one side. Rearranging:

4. Determine the degree.

   The degree is the exponent of the highest-order derivative ($\frac{d^3y}{dx^3}$) when the equation is a polynomial in derivatives.

   Look at each term involving $\frac{d^3y}{dx^3}$:

   • First term: $\left(\frac{d^2y}{dx^2}\right)^5 \cdot \frac{d^3y}{dx^3}$ → exponent 1 on the third derivative.
   • Third term: $\left(\frac{d^3y}{dx^3}\right)^2$ → exponent 2.
   • Fourth term: $-(x^2-1)\frac{d^3y}{dx^3}$ → exponent 1.

   The highest exponent on $\frac{d^3y}{dx^3}$ is 2. Hence the degree is $2$.

   Watch out

   A common mistake is to think the degree is 5 because of the first term’s power on the second derivative — but degree is defined by the power of the highest-order derivative only.

5. Sum the order and degree.

   Order = 3, Degree = 2 → Sum = $3 + 2 = 5$.

Concept: order = highest derivative present; degree = power of the highest-order derivative AFTER the equation is made free of radicals/fractional powers in the derivatives.

Equation: [1 - (dy/dx)^2]^(3/2) = k x (d^2y/dx^2).

Order: the highest derivative is d^2y/dx^2, so order = 2.

Degree: square both sides to clear the 3/2 power:

[1 - (dy/dx)^2]^3 = k^2 x^2 (d^2y/dx^2)^2.

Now the equation is polynomial in the derivatives, and the power of the highest-order derivative d^2y/dx^2 is 2. So degree = 2.

Sum = order + degree = 2 + 2 = 4.

ANSWER: B