1.1 Order and degree of a differential equation

Let us assume the differential equation in general form as $${f_1}(x,y){\left[ {\frac{{{d^m}y}}{{d{x^m}}}} \right]^{{n_1}}} + {f_2}(x,y){\left[ {\frac{{{d^{m - 1}}y}}{{d{x^{m - 1}}}}} \right]^{{n_2}}} + ... + {f_k}(x,y){\left[ {\frac{{dy}}{{dx}}} \right]^{{n_k}}} = 0\quad ...(1)$$

Order is the highest differential coefficient appearing in the differential equation i.e., in equation $(1)$, order is $m$.

Degree is the exponent of the highest differential coefficient. The exponent is to be checked after the differential equation is cleared of radicals or fractions i.e., when the differential equation is a polynomial in all the differential coefficients. As shown in equation $(1)$, the degree is ${n_1}$.

Question 1. Find the order and degree of the following differential equations:

(i) $\frac{{{d^2}y}}{{d{x^2}}} = {\left[ {y + {{\left( {\frac{{dy}}{{dx}}} \right)}^6}} \right]^{\frac{1}{2}}}$

(ii) $\frac{{{d^2}y}}{{d{x^2}}} = x\ln \left( {\frac{{dy}}{{dx}}} \right)$

(iii) ${e^{{y^{'''}}}} - x{y^{''}} - y = 0$

Solution:

(i) As the given differential equation is not a polynomial in all the differential coefficients, squaring both sides, we get $${\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} = \left[ {y + {{\left( {\frac{{dy}}{{dx}}} \right)}^6}} \right]$$ From the above equation, we can say that, order is $2$ and degree is $2$.

(ii) Order is $2$.

Since, the differential equation cannot be written as a polynomial in all differential coefficients, the degree is not defined.

(iii) The given differential equation can be written as $${e^{\frac{{{d^3}y}}{{d{x^3}}}}} - x\frac{{{d^2}y}}{{d{x^2}}} + y = 0$$

Order is $3$.

Since, the differential equation cannot be written as a polynomial in all differential coefficients, the degree is not defined.