How do you find the particular solution to #yy'e^x=0# that satisfies y(0)=4?
The differential equation:
is separable, so we can solve it by separating he variables and integrating:
So the particular solution we are seeking is:
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To find the particular solution to the differential equation (yy'  e^x = 0) that satisfies (y(0) = 4), follow these steps:

Separate variables to obtain: [ yy' = e^x ]

Integrate both sides with respect to (x): [ \int y , dy = \int e^x , dx ]

Integrate: [ \frac{1}{2} y^2 = e^x + C ]

Solve for (y): [ y^2 = 2(e^x + C) ]

Use the initial condition (y(0) = 4) to find (C): [ 4^2 = 2(e^0 + C) ] [ 16 = 2(1 + C) ] [ 8 = 1 + C ] [ C = 7 ]

Substitute (C) back into the equation: [ y^2 = 2(e^x + 7) ]

Take the square root of both sides to find (y): [ y = \pm \sqrt{2(e^x + 7)} ]
Given the initial condition (y(0) = 4), we choose the positive root: [ y = \sqrt{2(e^x + 7)} ]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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