# Differential Equations

Solve the Initial Value Problems (IVP) below:

$a)\;\;(\sin&space;x)y'&space;+&space;(\cos&space;x)y&space;=&space;0&space;;&space;\,y(\frac{7\pi}{6})&space;=&space;-2$

$b)\;\;y'&space;+&space;2(t+1)y^{2}&space;=&space;0;\,&space;y(0)&space;-&space;\frac{1}{8}$

${\color{Blue}&space;a)\;\;(\sin&space;x)y'&space;+&space;(\cos&space;x)y&space;=&space;0&space;;&space;\,y(\frac{7\pi}{6})&space;=&space;-2}$

##### Solution

This is an homogeneous differential equation and it is separable. First, we will divide everything by sin x

$\frac{\sin&space;x}{\sin&space;x}y'&space;+&space;\frac{\cos&space;x}{\sin&space;x}y&space;=&space;0$

$y'&space;=&space;-&space;\frac{\cos&space;x}{\sin&space;x}y&space;\;\;\;\;\;\;\;\;&space;[1]$

Now, we will make the following change:

$y'&space;=&space;\frac{dy}{dx}$

in the equation [1]  above. Then,

$\frac{dy}{dx}&space;=&space;-&space;\frac{\cos&space;x}{\sin&space;x}y&space;\;\;\;\;\;\;\;\;&space;[2]$

Working with this equation a little bit we get:

$\frac{dy}{y}&space;=&space;-&space;\frac{\cos&space;x}{\sin&space;x}&space;dx$

Integrating both sides we get:

$\int&space;\frac{1}{y}&space;dy&space;=&space;-&space;\int&space;\frac{\cos&space;x}{\sin&space;x}&space;dx$

$\ln&space;y&space;=-&space;ln|\sin&space;x|&space;+&space;c_{1}$

$\ln&space;y&space;=(ln|\sin&space;x|)^{-1}&space;+&space;c_{1}$

$\ln&space;y&space;=\frac{1}{ln|\sin&space;x|}&space;+&space;c_{1}$

$y&space;=&space;e^{\frac{1}{\ln&space;|&space;\sin&space;x|}&space;+&space;c_{1}}$

$y&space;=&space;e^{\frac{1}{\ln&space;|&space;\sin&space;x|}}&space;\cdot&space;e^{c_{1}}$

$y&space;=&space;ce^{\frac{1}{\ln&space;|&space;\sin&space;x|}}&space;\;\;\;\;\;\;\;\;&space;[3]$

where

$c&space;=&space;e^{c_{1}}$

Using the IVP value given in the problem we will have:

$y\left&space;(&space;\frac{7\pi}{6}&space;\right&space;)&space;=&space;-2&space;=&space;ce^{\frac{1}{\ln&space;|\sin&space;\left&space;(&space;\frac{7\pi}{6}&space;\right&space;)|}}&space;=&space;ce^{\frac{1}{\ln&space;\left&space;|&space;-\frac{1}{2}&space;\right&space;|}}&space;=&space;ce^{\frac{1}{\ln&space;\frac{1}{2}}}$

Making the necessary calculations, we will get:

$c&space;=&space;1.386294361&space;\approx&space;1.39$

Using this result in [3] above, we finally arrive at:

$\large&space;\mathit{y&space;=&space;1.39e^{\frac{1}{\ln&space;|&space;\sin&space;x|}}}$

Which is the equation we’ve been looking for.

${\color{Blue}&space;b)\;\;y'&space;+&space;2(t+1)y^{2}&space;=&space;0;\,&space;y(0)&space;-&space;\frac{1}{8}}$

##### Solution

This is also an Homogeneous Differential Equation and it is also separable. Repeating the same steps taken above to separate the variables, we will get the following equation:

$\frac{dy}{dt}&space;=&space;-2(t+1)y^{2}$

where

$y'&space;=&space;\frac{dy}{dt}$

then,

$\frac{1}{y^{2}}dy&space;=&space;-2(t+1)dt&space;\;\;\;\;\;\;\;\;&space;[4]$

Integrating, we get:

$\int&space;\frac{1}{y^{2}}dy&space;=&space;-&space;\int&space;2(t+1)dt&space;=&space;-\int&space;2t&space;dt&space;-&space;\int&space;dt$

$-\frac{1}{y}&space;=&space;-t^{2}&space;-&space;t&space;+&space;c&space;\Leftrightarrow&space;\frac{1}{y}&space;=&space;t^{2}&space;+&space;t&space;-&space;c$

Which implies that:

$y&space;=&space;\frac{1}{t^{2}&space;+&space;t&space;-c}&space;\;\;\;\;\;\;\;\;&space;[5]$

Using the IVP value in [5] above, we get:

$y(0)&space;=&space;-\frac{1}{8}&space;=&space;\frac{1}{t^{2}&space;+&space;t&space;-&space;c}&space;\Rightarrow&space;-8&space;=&space;t^{2}&space;+&space;t&space;-&space;c$

$c&space;=&space;t^{2}&space;+&space;t&space;+8&space;=&space;0&space;+&space;0&space;+&space;8&space;\Rightarrow&space;c&space;=&space;8&space;\;\;\;\;\;\;\;\;&space;[6]$

substituting [6] in [5] we finally have:

$\large&space;\mathit{y&space;=&space;\frac{1}{t^{2}&space;+&space;t&space;-&space;8}}$

Which is a particular solution for the given Differential Equation.

$\square$