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# separation of variables

Thanks

#### Solution Preview

1. The ODE is given as a first-order linear ODE. Here we have:
2x y' - ln(x^2) = 0

You have to use separation of variables to solve this ODE:

2x * y' = ln(x^2)
=> 2x * (dy/dx) = ln(x^2)

==> 2x*dy = ln(x^2) * dx
==> dy = ln(x^2)/2x * dx

We must now integrate both sides:

y = Integral[ln(x^2)/2x] *dx

This is not a complicated integral, since this is an ODE, I assume you know calculus, it's just integration by parts, twice:

You get, y = (1/8)*(ln(x^2))^2 + C, where C is some constant.

This is the general solution to the equation above.

2. Here, I let t = theta.

y' cos(t) = sin(t) + 2.5
Re-writing y' = dy/dt:

(dy/dt)cos(t) = sin(t) + 2.5

=> dy (cos(t)) = [sin(t) + 2.5]*dt

==> dy = [sin(t) + 2.5]/(cos(t)) * dt

Integrate both sides, by simplifying that sin[t]/cos[t] = tan[t], and 2.5/cos[t] = 2.5*sec[t]

==> y = Integral[tan[t] + 2.5*sec[t]] ...

#### Solution Summary

The method of integrating factors and method of separation of variables are encompassed.

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