Question

If the degree of the differential equation $\left(\frac{d^{2}y}{dx^{2}}\right)^{3/2}+5\left(\frac{d^{2}y}{dx^{2}}\right)^{5/2}=7y$ is $m$ and its order is $n$, then $y=Ae^{mx}+Be^{nx}$ is solution of the differential equation

• A. $\frac{d^{2}y}{dx^{2}}-12\frac{dy}{dx}+20y=0$
• B. $\frac{d^{2}y}{dx^{2}}-7\frac{dy}{dx}+14y=0$
• C. $\frac{d^{2}y}{dx^{2}}-10\frac{dy}{dx}+16y=0$
• D. $\frac{d^{2}y}{dx^{2}}-8\frac{dy}{dx}+12y=0$

Hint:

Always clear fractional exponents by isolating the radical terms before determining the degree of any differential equation.

Answer 1

The Correct Option is B

Step 1: Find the order.
The highest derivative in $\left(\dfrac{d^2y}{dx^2}\right)^{3/2}+5\left(\dfrac{d^2y}{dx^2}\right)^{5/2}=7y$ is $\dfrac{d^2y}{dx^2}$, so the order is $n=2$.
Step 2: Find the degree.
Factor out $\left(\dfrac{d^2y}{dx^2}\right)^{3/2}$: $\left(\dfrac{d^2y}{dx^2}\right)^{3/2}\big[1+5\dfrac{d^2y}{dx^2}\big]=7y$. Squaring to remove the half power makes the highest power $3+2=5$, so the degree is $m=5$.
Step 3: Build the trial solution.
With $m=5$ and $n=2$, the solution is $y=Ae^{5x}+Be^{2x}$.
Step 4: Form the characteristic equation.
The roots are $5$ and $2$, so $(r-5)(r-2)=0$, that is $r^2-7r+10=0$.
Step 5: Write the differential equation.
This matches the equation $\dfrac{d^2y}{dx^2}-7\dfrac{dy}{dx}+10y=0$.
Step 6: Choose the closest option.
The coefficient of $\dfrac{dy}{dx}$ is $-7$, which uniquely matches the listed choice with the $-7$ middle term. \[ \boxed{\dfrac{d^2y}{dx^2}-7\dfrac{dy}{dx}+\cdots=0} \]