solve each differential equation, applying the initial condition when provided. dy/dx=e^x/2y and dy/dx=3y y(0)=5

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dy/dx=e^x/2y

2y dy = e^x dx
y^2 = e^x + c

dy/dx = 3y
dy/y = 3 dx
ln y = 3x+c
y = ce^(3x)
y(0)=5, so c=5
y = 5e^(3x)

To solve each differential equation and apply the initial condition, we will use the method of separation of variables.

1. dy/dx = (e^x) / (2y)
First, separate the variables by multiplying both sides of the equation by 2y and dividing by e^x:
2y * dy = (e^x) * dx

Next, integrate both sides of the equation:
∫ 2y * dy = ∫ e^x * dx

To integrate 2y dy, we divide by 2 and integrate y with respect to y:
y^2 = ∫ e^x * dx

Integrating e^x is straightforward:
y^2 = e^x + C

Now, applying the initial condition:
If y(0) = 5, substitute x = 0 and y = 5 into the equation:
5^2 = e^0 + C
25 = 1 + C

So, C = 24.
Therefore, the solution to the differential equation with the given initial condition is:
y^2 = e^x + 24

2. dy/dx = 3y
Again, separate the variables:
dy / y = 3 dx

Integrate both sides:
∫ dy / y = ∫ 3 dx

The integral of dy / y is the natural logarithm of the absolute value of y:
ln|y| = 3x + C

Applying the initial condition:
If y(0) = 5,
ln|5| = 3(0) + C
ln(5) = C

Therefore, C = ln(5).
The solution to the differential equation with the given initial condition is:
ln|y| = 3x + ln(5)

Note: In both cases, we end up with general solutions to the differential equations. If you need specific solutions for particular values of x, you would need additional information like boundary conditions.