Question

Degree of the differential equation $(y^{\prime \prime }{)}^2-\sqrt{y^{\prime }}=y^3$ is:

• A. $2$
• B. $3$
• C. $4$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $4$

The degree of a differential equation is the degree of highest order derivative term, after it has made rational and integral as far as the derivatives are concerned in the equation.

Now for the given differential equation $(y^{\prime \prime }{)}^2-\sqrt{y^{\prime }}=y^3$ can be written as

$(y^{\prime \prime }{)}^2-y^3=\sqrt{y^{\prime }}$

or, $\left\{\right(y^{\prime \prime }{)}^2-y^3{\}}^2=y^{\prime }$

From this equation it is clear that the degree of highest order derivative $y^{\prime \prime }$ is $4$.

So the degree of the differential equation is $4$.