# How do you find the particular solution to #(du)/(dv)=uvsinv^2# that satisfies u(0)=1?

We can find the general solution by separating the variables:

so

The particular solution we look for is then:

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To find the particular solution to (\frac{du}{dv} = u \cdot v \cdot \sin(v^2)) that satisfies (u(0) = 1), we can separate variables and integrate both sides.

Starting with (\frac{du}{dv} = u \cdot v \cdot \sin(v^2)), we separate variables to get:

[\frac{du}{u} = v \cdot \sin(v^2) , dv]

Next, integrate both sides:

[\int \frac{du}{u} = \int v \cdot \sin(v^2) , dv]

The left side integrates to (\ln|u|), and the right side can be solved with a substitution. Let (w = v^2), then (dw = 2v , dv), and the integral becomes:

[\frac{1}{2} \int \sin(w) , dw]

Integrating (\sin(w)) gives (-\cos(w)), so the integral becomes:

[-\frac{1}{2} \cos(w) + C]

Substitute back (w = v^2) and simplify:

[-\frac{1}{2} \cos(v^2) + C]

Now we have:

[\ln|u| = -\frac{1}{2} \cos(v^2) + C]

Since (u(0) = 1), we can substitute (u = 1) and (v = 0) into the equation:

[\ln(1) = -\frac{1}{2} \cos(0) + C]

This simplifies to (0 = -\frac{1}{2} + C), so (C = \frac{1}{2}).

Thus, the particular solution is:

[\ln|u| = -\frac{1}{2} \cos(v^2) + \frac{1}{2}]

Exponentiate both sides to solve for (u):

[|u| = e^{-\frac{1}{2} \cos(v^2) + \frac{1}{2}}]

However, since we are given (u(0) = 1), we take the positive branch:

[u = e^{-\frac{1}{2} \cos(v^2) + \frac{1}{2}}]

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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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