Question

If $m$ and $n$ are respectively the order and degree of the differential equation $2x^2\dfrac{d^2y}{dx^2}-3\dfrac{dy}{dx}+y=0$, then $m+n=$

Hint:

Order means the highest derivative present, while degree means the power of that highest derivative after writing the equation in polynomial form in derivatives.

Answer 1

Step 1: Recall the definition of order of a differential equation.
The order of a differential equation is the highest order derivative present in the equation.

Step 2: Identify the highest derivative.
The equation given is \[ 2x^2\frac{d^2y}{dx^2}-3\frac{dy}{dx}+y=0 \] The highest derivative present is \[ \frac{d^2y}{dx^2} \] Thus the order of the differential equation is \[ m=2 \]
Step 3: Recall the definition of degree.
The degree of a differential equation is the power of the highest order derivative when the equation is polynomial in derivatives.

Step 4: Determine the degree.
The highest derivative \[ \frac{d^2y}{dx^2} \] appears with power $1$. Thus the degree of the equation is \[ n=1 \]
Step 5: Compute the required sum.
\[ m+n=2+1 \] \[ m+n=3 \]
Step 6: Conclusion.
Hence the sum of order and degree of the differential equation is $3$.
Final Answer: $\boxed{3}$