Question

Given the differential equation dy/dx = x^4-y^4/(x^2+y^2)xy, the degree of the differential vequation is

Answer 1

The degree of the differential equation = 1

Given :

The differential equation

$\displaystyle \sf{ \frac{dy}{dx} = \frac{ {x}^{4} - {y}^{4} }{( {x}^{2} + {y}^{2} )xy} }$

To find :

The degree of the differential equation

Concept :

Degree of a differential equation :

The degree of a differential equation is the degree of the highest derivative occuring in it after the equation has been expressed in a form free from radicals and fractions as far as the derivatives are concerned

Solution :

Step 1 of 2 :

Write down the given differential equation

The given differential equation is

$\displaystyle \sf{ \frac{dy}{dx} = \frac{ {x}^{4} - {y}^{4} }{( {x}^{2} + {y}^{2} )xy} }$

Step 2 of 2 :

Find degree of the differential equation

For the differential equation

$\displaystyle \sf{ \frac{dy}{dx} = \frac{ {x}^{4} - {y}^{4} }{( {x}^{2} + {y}^{2} )xy} }$

The highest derivative occuring in the differential equation after the equation has been expressed in a form free from radicals and fractions as far as the derivatives are concerned is 1

Hence degree of the differential equation = 1

Correct question : Given the differential equation $\displaystyle \sf{ \frac{dy}{dx} = \frac{ {x}^{4} - {y}^{4} }{( {x}^{2} + {y}^{2} )xy} }$ the degree of the differential equation is

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ M+N(dy/dx)=0 where M and N are function of

https://brainly.in/question/38173299

2. This type of equation is of the form dy/dx=f1(x,y)/f2(x,y)

https://brainly.in/question/38173619