What is a solution to the differential equation #y'-y=5#?

Answer 1

#y= C e^x - 5#

separate it!

#y' - y = 5#
#y' = y + 5#
#1/(y+5) y' = 1#
#int 1/(y+5) y' \ dx=int \dx#
#int d/dx (int 1/(y+5) \dy) \ dx=int \dx#
#int 1/(y+5) \dy=int \dx#
#ln(y+5) =x + C#
#y+5 =e^(x + C) = C e^x#
#y= C e^x - 5#
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Answer 2

The solution to the differential equation ( y' - y = 5 ) can be found by first solving the associated homogeneous equation, which is ( y' - y = 0 ). The solution to this homogeneous equation is ( y_h = Ce^x ), where ( C ) is an arbitrary constant.

To find the particular solution to the non-homogeneous equation ( y' - y = 5 ), we can use the method of undetermined coefficients. Since the right-hand side is a constant, we can assume a particular solution of the form ( y_p = A ), where ( A ) is a constant.

Substituting ( y_p = A ) into the original differential equation, we get: [ 0 - A = 5 ] [ A = -5 ]

Therefore, the particular solution to the non-homogeneous equation is ( y_p = -5 ).

The general solution to the differential equation ( y' - y = 5 ) is the sum of the homogeneous and particular solutions: [ y = y_h + y_p = Ce^x - 5 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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