# How do you solve the separable differential equation #dy/dx=(cosx)e^(y+sinx)#?

# y = ln(1/(A-e^sinx)) # is the General Solution

We have:

Which is a First Order Separable Differential Equation, which we can rewrite as:

We can then "separate the variables" to get:

Which we can directly (and easily) integrate to get:

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Now we can integrate both sides:

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To solve the separable differential equation dy/dx = (cosx)e^(y+sinx), we can separate the variables and integrate both sides with respect to their respective variables. First, we divide both sides by (cosx)e^(y+sinx) to isolate dy/dx. This gives us:

dy / (cosx)e^(y+sinx) = dx

Next, we integrate both sides. The integral of dy / (cosx)e^(y+sinx) with respect to y involves a substitution. Let u = y + sinx, then du = dy. The integral becomes:

∫ (1 / (cosx)e^u) du

This integral can be simplified using trigonometric identities. Rewrite (1 / cosx) as secx, and the integral becomes:

∫ (secx / e^u) du

Now, we can integrate ∫ (secx / e^u) du. This integral involves integration by parts. Let dv = secx du and u = e^u. Then, differentiate u to find du, and integrate dv to find v. We have:

dv = secx du du = e^u du

Integrating dv gives v = ln|secx + tanx| + C, where C is the constant of integration.

Now, we can apply integration by parts to find the integral of ∫ (secx / e^u) du. This gives us:

∫ (secx / e^u) du = ln|secx + tanx| + C

Now, we substitute back u = y + sinx:

ln|secx + tanx| = y + sinx + C

This is the solution to the differential equation dy/dx = (cosx)e^(y+sinx).

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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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