What is the Implicit answer for this differential equation?

$\frac{dy}{dx}={y}^{2}-4$

Milton Anderson
2022-09-14
Answered

I need help with Differential equations.

What is the Implicit answer for this differential equation?

$\frac{dy}{dx}={y}^{2}-4$

What is the Implicit answer for this differential equation?

$\frac{dy}{dx}={y}^{2}-4$

You can still ask an expert for help

Leon Webster

Answered 2022-09-15
Author has **17** answers

Step 1

Here is how you solve equations like yours. You are trying to solve $\frac{dy}{dx}=f(y)$. Rewrite it as $\frac{dy}{f(y)}=dx$, integrate on both sides $\int \frac{1}{f(y)}dy=\int dx=x+c$ and invert to get expression for y(x). If you have conditions on the value of y(x) as some particular x, like $x=0$, then you can determine c as well.

Step 2

Applying this to your example. You have $\int \frac{1}{{y}^{2}-4}dy=x+c$. The integral is $\frac{1}{4}\mathrm{log}\left(\frac{2-y}{2+y}\right)$ so that $\frac{2-y}{2+y}=\mathrm{exp}(4x+4c)$. From here solving for y is straightforward.

Here is how you solve equations like yours. You are trying to solve $\frac{dy}{dx}=f(y)$. Rewrite it as $\frac{dy}{f(y)}=dx$, integrate on both sides $\int \frac{1}{f(y)}dy=\int dx=x+c$ and invert to get expression for y(x). If you have conditions on the value of y(x) as some particular x, like $x=0$, then you can determine c as well.

Step 2

Applying this to your example. You have $\int \frac{1}{{y}^{2}-4}dy=x+c$. The integral is $\frac{1}{4}\mathrm{log}\left(\frac{2-y}{2+y}\right)$ so that $\frac{2-y}{2+y}=\mathrm{exp}(4x+4c)$. From here solving for y is straightforward.

Modelfino0g

Answered 2022-09-16
Author has **1** answers

Step 1

$$\frac{dy}{dx}={y}^{2}-4\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{dy}{{y}^{2}-4}=dx$$

Then integrate each side. You can easily use partial fractions on the left hand side.:

$$\int \frac{dy}{{y}^{2}-4}=\int \frac{dy}{(y-2)(y+2)}=\int \frac{A}{y-2}+\frac{B}{y+2}$$

$Ay+2A+By-2B=1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(A+B)y+2(A-B)=1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(A+B=0,\phantom{\rule{thickmathspace}{0ex}}2(A-B)=1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(A=-B\wedge -4B=1).$

So $A=\frac{1}{4},B=-\frac{1}{4}$

That gives us

$$\frac{1}{2}\int \frac{1}{y-2}-\frac{1}{y+2}dy=\int dx+c$$

$$\begin{array}{}\text{(1)}& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{1}{4}\mathrm{ln}\left(\frac{y-2}{y+2}\right)=x+C\end{array}$$

Step 2

Given your last comment, we can continue as follows from (1):

$$\begin{array}{}\text{(2)}& \mathrm{ln}\left(\frac{y-2}{y+2}\right)=4x+{C}^{\ast}\end{array}$$

Raising each side as a power of e gives us:

$$\mathrm{ln}\left(\frac{y-2}{y+2}\right)=4x+C\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{e}^{\mathrm{ln}\left(\frac{y-2}{y+2}\right)}={e}^{4x+C}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{y-2}{y+2}={e}^{4x}\underset{{C}^{\prime}}{\underset{\u23df}{{e}^{C}}}$$

So $\begin{array}{}\text{(3)}& \frac{y-2}{y+2}={C}^{\prime}{e}^{4x}\end{array}$

$$\frac{dy}{dx}={y}^{2}-4\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{dy}{{y}^{2}-4}=dx$$

Then integrate each side. You can easily use partial fractions on the left hand side.:

$$\int \frac{dy}{{y}^{2}-4}=\int \frac{dy}{(y-2)(y+2)}=\int \frac{A}{y-2}+\frac{B}{y+2}$$

$Ay+2A+By-2B=1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(A+B)y+2(A-B)=1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(A+B=0,\phantom{\rule{thickmathspace}{0ex}}2(A-B)=1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}(A=-B\wedge -4B=1).$

So $A=\frac{1}{4},B=-\frac{1}{4}$

That gives us

$$\frac{1}{2}\int \frac{1}{y-2}-\frac{1}{y+2}dy=\int dx+c$$

$$\begin{array}{}\text{(1)}& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{1}{4}\mathrm{ln}\left(\frac{y-2}{y+2}\right)=x+C\end{array}$$

Step 2

Given your last comment, we can continue as follows from (1):

$$\begin{array}{}\text{(2)}& \mathrm{ln}\left(\frac{y-2}{y+2}\right)=4x+{C}^{\ast}\end{array}$$

Raising each side as a power of e gives us:

$$\mathrm{ln}\left(\frac{y-2}{y+2}\right)=4x+C\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{e}^{\mathrm{ln}\left(\frac{y-2}{y+2}\right)}={e}^{4x+C}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{y-2}{y+2}={e}^{4x}\underset{{C}^{\prime}}{\underset{\u23df}{{e}^{C}}}$$

So $\begin{array}{}\text{(3)}& \frac{y-2}{y+2}={C}^{\prime}{e}^{4x}\end{array}$

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as an exact differential equation and I know it's exact because I solve the equaliy

$$\frac{\mathrm{\partial}(2x+y)}{\mathrm{\partial}y}=1$$

and

$$\frac{\mathrm{\partial}(x-2y)}{\mathrm{\partial}x}=1$$

so following the steps to solve this kind of equations i have:

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and

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to be honest I have many doubts what are the next steps so if you can guide me I'll apreciate

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