Question

What is the degree of first order differential equation, given by 15=(x
cosx.x2+ysinx)
O a) 1.5
O b) 1
O c) 3
O d) 0.5​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

the degree of first order differential equation is 1.

Given:

15=(x cosx.x2+ysinx)

To find:

The degree of first order differential equation

Solution:

15 = x³cosx + ysinx

ysinx = x³cosx - 15

y = (x³cosx - 15)/ sinx

y = x³ tanx - 15cosecx

dy/dx = - 3x²sec²x + cotx.cosecx

The highest order derivative of a differential equation is defined as its degree, which is expressed as a power. An illustration of a second-degree, third-order differential equation is (f′′′)2 + (f′′)4 + f = x.

Here, the largest differentiation is 1, so it's degree will also be 1 as there is power 1 on dy/dx.

So, the degree of first order differential equation is 1.

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