Step 1: Understanding the Concept:
The order of a differential equation is the order of the highest derivative present.
The degree is the power of the highest order derivative when the equation is expressed as a polynomial in all its derivatives. This means we must eliminate all fractional powers (radicals) involving derivatives.
Step 2: Key Formula or Approach:
Given equation: $\left(\frac{d^2y}{dx^2}\right)^{1/2} = \left(\frac{dy}{dx}\right)^{1/5} - 5$.
Identify highest derivative for order.
Algebraically manipulate the equation to remove radicals. If the equation is $A^{1/p} = B^{1/q} + C$, isolating terms and raising to LCM of powers isn't always straight forward. Let's substitute variables to simplify algebra.
Step 3: Detailed Explanation:
Let $y'' = \frac{d^2y}{dx^2}$ and $y' = \frac{dy}{dx}$.
The equation is $\sqrt{y''} = (y')^{1/5} - 5$.
The highest order derivative is $y''$, which is a second-order derivative.
Therefore, the Order = 2.
To find the degree, we need to make it a polynomial in $y''$ and $y'$.
Let's isolate $y'$ first, as it has a higher root.
$\sqrt{y''} + 5 = (y')^{1/5}$
Raise both sides to the power of 5:
\[ (\sqrt{y''} + 5)^5 = y' \]
Now, expand the left side using the Binomial Theorem:
\[ (\sqrt{y''})^5 + \binom{5}{1}(\sqrt{y''})^4(5) + \binom{5}{2}(\sqrt{y''})^3(5^2) + \binom{5}{3}(\sqrt{y''})^2(5^3) + \binom{5}{4}(\sqrt{y''})^1(5^4) + 5^5 = y' \]
\[ (y'')^{5/2} + 25(y'')^2 + 250(y'')^{3/2} + 1250y'' + 3125(y'')^{1/2} + 3125 = y' \]
We still have fractional powers of $y''$: $1/2, 3/2, 5/2$. Let's gather all terms with fractional powers on one side and integer powers on the other.
\[ (y'')^{5/2} + 250(y'')^{3/2} + 3125(y'')^{1/2} = y' - 25(y'')^2 - 1250y'' - 3125 \]
Factor out $(y'')^{1/2}$ on the left side:
\[ (y'')^{1/2} [ (y'')^2 + 250y'' + 3125 ] = y' - 25(y'')^2 - 1250y'' - 3125 \]
Now, to eliminate the fractional power $(y'')^{1/2}$, we square both sides:
\[ \left( (y'')^{1/2} [ (y'')^2 + 250y'' + 3125 ] \right)^2 = \left( y' - 25(y'')^2 - 1250y'' - 3125 \right)^2 \]
\[ y'' [ (y'')^2 + 250y'' + 3125 ]^2 = \left( y' - 25(y'')^2 - 1250y'' - 3125 \right)^2 \]
This equation is now a polynomial in $y''$ and $y'$.
The degree is the highest exponent of the highest order derivative, $y''$.
Let's find the term with the highest power of $y''$ on both sides.
On the left side: The highest power term comes from multiplying $y''$ with the square of the highest power term in the bracket.
LHS highest power term $\approx y'' \cdot ((y'')^2)^2 = y'' \cdot (y'')^4 = (y'')^5$.
On the right side: The highest power term comes from squaring the highest power term of $y''$.
RHS highest power term $\approx (-25(y'')^2)^2 = 625(y'')^4$.
Comparing both sides, the overall highest power of $y''$ in the entire expanded equation is 5.
Therefore, the Degree = 5.
The sum of degree and order is $5 + 2 = 7$.
Step 4: Final Answer:
The sum of the degree and order is $7$.