Solve the following ode y' + tan(5x).y = 0 with initial condition y(0) = 4

Neetesh Kumar | August 29, 2024

This is the solution to Myopenmath Math2A Differential Equation homework question number 2.2.13

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Step-by-Step-Solution: Given $y' + tan(5x).y= 0$

We can solve the above-given IVP using the separation of variables method.

$\frac{dy}{y} = -tan(5x)dx$

Now, after integrating both sides

$\int \frac{dy}{y} = \int -tan(5x)dx$

$ln|y| = -\frac{1}{5}.ln|sec(5x)| + ln|c|$

$y = c(sec(5x))^{\frac{-1}{5}}$

Using the initial condition y(0) = 4 we can find the value of c

$4 = c$

Then, we can write the general solution of the above-given IVP as

Answer: $y = 4(cos(5x))^{\frac{1}{5}}$

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