degree of the differential equation -1
What is the degree of the differential equation y = 1 + d x d y + 2 ! 1 . ( d x d y ) 2 + 3 ! 1 . ( d x d y ) 3 + 4 ! 1 . ( d x d y ) 4 + . . . . . .
Note : If degree is not defined, write 1 7 2 9 as your answer.
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The answer is 1.
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4 solutions
the degree is applied to the polynomial function not logarithmic ones. y must be in the form a 0 x + a 1 x + a 2 x 2 + . . . . + a n x n , thus, it's of the n-th degree.
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Degree is applied when the differential equation is 'polynomial in derivatives' . Not nessecary that whole function must be polynomial. Above solution is correct as it is polynomial in derivatives.
First differentiate it once and substitute the value of y from first equation in second and the equation becomes Y'=Y"Y. Hence the degree is 1
In general, the degree is the power of highest ordered derivative occurring in the equation and order is highest ordered derivative occurring in the equation. So, in given eq. degree is 1
according to what you said, I want to edit the definition of the degree as it is the highest power of the derivative. so, the given equation is of order 1, but its degree is undefined, as we don't know the highest power of the derivative.
dy/dx power mean whole power. so the degree of differential equation is the degree of differential.
we see that y= e^(dy/dx) that results: ln(y) = dy/dx, hence degree is 1