I try to solve this equation by finding

How am I supposed to solve it?

Ali Marshall
2022-05-01
Answered

I try to solve this equation by finding

How am I supposed to solve it?

You can still ask an expert for help

Jonas Dickerson

Answered 2022-05-02
Author has **22** answers

Step 1

There is a method to solve equations like this one that uses a clever substitution. The idea is to use substitution to remove the y' term. Whenever you have an equation in the form of

$y{}^{\u2033}+f\left(x\right){y}^{\prime}+g\left(x\right)y=0$, you substitute $y=\mathrm{exp}(-\int \frac{f\left(x\right)}{2},dx)\mu$. Divide through by $x}^{2$ to get

$y{}^{\u2033}+\underset{=f\left(x\right)}{\underset{\u23df}{\frac{3x+4}{x}}}{y}^{\prime}+\underset{=g\left(x\right)}{\underset{\u23df}{\frac{2({x}^{2}+3x+1)}{{x}^{2}}}}y=0.$

Then substitute in $y=\mathrm{exp}(-\int \frac{3x+4}{2x},dx)\mu =\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}\mu .$

Therefore,

$\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}\mu {}^{\u2033}-\frac{{e}^{-3\frac{x}{2}}}{4{x}^{2}}\mu =0$.

Step 2

Divide through by $\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}$ to get

$\mu {}^{\u2033}-\frac{1}{4}\mu =0.$

Using standard methods in ODEs, we get

$\mu ={c}_{1}{e}^{-\frac{x}{2}}+{c}_{2}{e}^{\frac{x}{2}}$.

Therefore, $y=\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}\mu =\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}({c}_{1}{e}^{-\frac{x}{2}}+{c}_{2}{e}^{\frac{x}{2}})$.

asked 2021-01-02

Find

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Given population doubles in 20 minutes, what is intrinsic growth rate r?

Attempt: Given population doubles, using exponential growth rate we have$\frac{dN}{dt}=2N$ so $N\left(t\right)={N}_{0}{e}^{2t}$ therefore $r=2$ , but I have a feeling this is wrong since 20 minutes should be used somewhere around here.

Given population doubles in 20 minutes, what is intrinsic growth rate r?

Attempt: Given population doubles, using exponential growth rate we have

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$$\begin{array}{rl}{x}_{1}^{\prime}& =2{x}_{1}+3{x}_{2},\\ {x}_{2}^{\prime}& ={x}_{1}-{x}_{2}.\end{array}$$

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The equation is$f}^{{}^{\prime}}\left(x\right)=\gamma \frac{f\left(x\right)+{f}^{2}\left(x\right)}{\mathrm{log}\left(\frac{f\left(x\right)}{1+f\left(x\right)}\right)$

with the initial condition f(0), where$x\ge \text{}\text{and}\text{}f\left(0\right)\ge 0$ .

The solution is

$f\left(x\right)=\frac{1}{-1+\mathrm{exp}\sqrt{2\gamma x+{\mathrm{log}\left(\frac{f\left(0\right)+1}{f\left(0\right)}\right)}^{2}}}$

I think the ODE can be solved by separating the variables

${\int}_{0}^{f\left(x\right)}\frac{\mathrm{log}\left(\frac{f\left(z\right)}{1+f\left(z\right)}\right)}{f\left(z\right)+{f}^{2}\left(z\right)}df={\int}_{0}^{x}\gamma dz$

The right-hand size is easy. I do not know how to solve the integration of the left-hand side.

The equation is

with the initial condition f(0), where

The solution is

I think the ODE can be solved by separating the variables

The right-hand size is easy. I do not know how to solve the integration of the left-hand side.

asked 2022-04-24

Infer boundedness from differential inequality

$\frac{dx}{dt}\le x{\left(t\right)}^{2}+y\left(t\right)$ ?