Find the solution of the differential equation that satisfies the given initial conditions?

(Just give me tips, you don't have to do the whole thing) a. $y y ' = x \setminus \sin x$ and $y \left(0\right) = - 1$ b. $y ' \setminus \tan x = y + 8$, $y \left(\setminus \frac{\pi}{3}\right) = 8$ and $0 \setminus < x \setminus < \setminus \frac{\pi}{2}$ c. $x y ' = y + {x}^{2} \setminus \sin x$ and $y \left(\setminus \pi\right) = 0$

Apr 24, 2018

$\left(a\right) \setminus {y}^{2} = 2 \sin x - 2 x \cos x + 1$
$\left(b\right) \setminus y = \frac{32 \sqrt{3}}{3} \sin x - 8$
$\left(c\right) \setminus y = - x \cos x - x$

Explanation:

Part (a):

$y y ' = x \sin x$ with $y \left(0\right) = - 1$

This is separable, so by "separating the variables" , we get:

$\int \setminus y \setminus \mathrm{dy} = \int \setminus x \sin x \setminus \mathrm{dx}$ ..... [A]

To integrate the RHS we apply IBP

Let  { (u,=x, => (du)/dx,=1), ((dv)/dx,=sinx, => v,=-cosx ) :}

Then plugging into the IBP formula:

$\int \setminus u \frac{\mathrm{dv}}{\mathrm{dx}} \setminus \mathrm{dx} = u v - \int \setminus v \frac{\mathrm{dv}}{\mathrm{dx}} \setminus \mathrm{dx}$

So we have:

$\int \setminus x \sin x \setminus \mathrm{dx} = - x \cos x + \int \setminus \cos x \setminus \mathrm{dx} = \sin x - x \cos x$

Thus if we integrate [A] we get:

$\frac{1}{2} {y}^{2} = \sin x - x \cos x + C$

Applying the initial conditions:

$y \left(0\right) = - 1 \implies \frac{1}{2} = 0 - 0 + C$

Thus we gain the particular solution

$\frac{1}{2} {y}^{2} = \sin x - x \cos x + \frac{1}{2}$

$\therefore {y}^{2} = 2 \sin x - 2 x \cos x + 1$

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Part (b):

$y ' \tan x = y + 8$ with $y \left(\frac{\pi}{3}\right) = 8$ and $0 < x < \frac{\pi}{2}$

This is also separable:

$\frac{1}{y + 8} y ' = \cot x$

We can "separate the variables" to get:

$\int \setminus \frac{1}{y + 8} \setminus \mathrm{dy} = \int \setminus \cot x \setminus \mathrm{dx}$

Which is trivial to integrate:

$\ln | y + 8 | = \ln | \sin x | + C$

Applying the initial conditions:

$y \left(\frac{\pi}{3}\right) = 8 \implies \ln | 8 + 8 | = \ln | \sin \left(\frac{\pi}{3}\right) | + C$
$\therefore \ln 16 = \ln \left(\frac{\sqrt{3}}{2}\right) + C \implies C = \ln \left(\frac{32 \sqrt{3}}{3}\right)$

Thus we gain the particular solution:

$\ln | y + 8 | = \ln | \sin x | + \ln \left(\frac{32 \sqrt{3}}{3}\right)$
$\therefore y + 8 = \frac{32 \sqrt{3}}{3} \sin x$
$\therefore y = \frac{32 \sqrt{3}}{3} \sin x - 8$

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Part (c):

$x y ' = y + {x}^{2} \sin x$ with $y \left(\pi\right) = 0$

We can write as:

$y ' - \frac{y}{x} = x \sin x$

Which is of the form:

$\frac{\mathrm{dy}}{\mathrm{dx}} + P \left(x\right) y = Q \left(x\right)$

So, we can construct an Integrating Factor:

 I = exp(int \ P(x) \ dx
$\setminus \setminus = \exp \left(\int \setminus - \frac{1}{x} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(- \ln \left(x\right)\right)$
$\setminus \setminus = \frac{1}{x}$

And if we multiply the DE by this Integrating Factor, $I$, we will have a perfect product differential;

$\frac{1}{x} \setminus y ' - \frac{1}{x} ^ 2 \setminus y = \frac{1}{x} \setminus x \sin x$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{1}{x} \setminus y\right) = \sin x$

This is now separable, so we can "separate the variables" to get:

$\frac{y}{x} = \int \setminus \sin x \setminus \mathrm{dx}$

Which is trivial to integrate:

$\frac{y}{x} = - \cos x + C$

Applying the initial conditions:

$y \left(\pi\right) = 0 \implies 0 = - \cos \pi + C \implies C = - 1$

Thus we gain the particular solution:

$\frac{y}{x} = - \cos x - 1$

$\therefore y = - x \cos x - x$