# How to you find the general solution of #4yy'-3e^x=0#?

Changing the notation from Lagrange's notation to Leibniz's notation we have:

This is a First Order sepaable DE, and "seperating the variables" give is

Integrating gives us:

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To find the general solution of the differential equation (4yy' - 3e^x = 0), you can use the method of separation of variables. Rearrange the equation to isolate (y) and (y') terms. Then integrate both sides with respect to (x) to solve for (y). The general solution will contain an arbitrary constant.

Starting with the given equation (4yy' - 3e^x = 0), we rearrange to isolate the (y') term:

[4yy' = 3e^x]

[y' = \frac{3e^x}{4y}]

Now, separate the variables by multiplying both sides by (dx) and dividing both sides by (y):

[\frac{dy}{dx} = \frac{3e^x}{4y}]

[4y , dy = 3e^x , dx]

Next, integrate both sides:

[\int 4y , dy = \int 3e^x , dx]

[2y^2 = 3e^x + C]

Finally, solve for (y):

[y^2 = \frac{3}{2} e^x + \frac{C}{2}]

[y = \pm \sqrt{\frac{3}{2} e^x + \frac{C}{2}}]

So, the general solution of the differential equation is:

[y = \pm \sqrt{\frac{3}{2} e^x + \frac{C}{2}}]

Where (C) is the constant of integration.

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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.

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