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