Question

Use the methods of this section to solve the following differential equations. Compare computer solutions and reconcile differences.

$\mathit{y}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{2}}\mathit{c}\mathit{o}\mathit{t}\mathit{x}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$

Short answer

Answer

The general solution of the differential equation is $y=\sqrt{-{\cos }^{2}xci{n}^{2}x+C{\sin }^{4}x}$

Step-by-step solution

Given information

The given differential equation is$yy\text{'}-2y^{2}cotx=\sin x\cos x$.

Definition of differential equation

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself to its derivatives of various orders.

Solve the differential equation

Rewrite the differential equation as,

$y\text{'}-2ycotx=\frac{\sin 2x}{2y}$

This equation is in the form of Bernoulli equation. So we can solve it by assuming $z=y^2$. Then, $z\text{'}=2yy\text{'}$.

$$\begin{gathered} 2yy\text{'}-4y^{2}co{t}^{2}cotx=\sin 2x \\ z\text{'}-4zcotx=\sin 2x \end{gathered}$$

Now it becomes a first order linear differential equation as,

$$\begin{gathered} I=-4\int cotxdx \\ =-4In\left(\sin x\right) \\ {e}^I={\sin }^{-4}x \end{gathered}$$

Solve the equation further as,

$$\begin{gathered} z{e}^I=\left(\sin 2x\right){\sin }^{4}xdx \\ =-co{t}^{2}x+C \\ z=-{\cos }^{2}x{\sin }^{2}x+C{\sin }^{4}x \end{gathered}$$

Finally, substitute $z=y^2$. So, the solution is $y=\sqrt{-{\cos }^{2}x{\sin }^{2}x+C{\sin }^{4}x}$.