What is a solution to the differential equation $\frac{dy}{dx}=1+2xy$?

Aubrie Aguilar
2022-09-22
Answered

What is a solution to the differential equation $\frac{dy}{dx}=1+2xy$?

You can still ask an expert for help

Amiya Watkins

Answered 2022-09-23
Author has **6** answers

$\frac{dy}{dx}=1+2xy$

$\therefore \frac{dy}{dx}-2xy=1$ ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

$I={e}^{\int P\left(x\right)dx}$

$={e}^{\int -2xdx}$

$={e}^{-{x}^{2}}$

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

$\frac{dy}{dx}-2xy=1$

$\therefore {e}^{-{x}^{2}}\frac{dy}{dx}-2xy{e}^{-{x}^{2}}=1\cdot {e}^{-{x}^{2}}$

$\therefore \frac{d}{dx}\left(y{e}^{-{x}^{2}}\right)={e}^{-{x}^{2}}$

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

$y{e}^{-{x}^{2}}=\int {e}^{-{x}^{2}}dx$

The RHS integral does not have an elementary form, but we can use the definition of the Error Function :

$\text{erf}\left(x\right)=\frac{2}{\sqrt{\pi}}{\int}_{0}^{x}{e}^{-{t}^{2}}dt$

Which gives us:

$y{e}^{-{x}^{2}}=\frac{\sqrt{\pi}}{2}\text{erf}\left(x\right)+A$

$y=\frac{\sqrt{\pi}}{2}{e}^{{x}^{2}}\text{erf}\left(x\right)+A{e}^{{x}^{2}}$

$\therefore \frac{dy}{dx}-2xy=1$ ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

$I={e}^{\int P\left(x\right)dx}$

$={e}^{\int -2xdx}$

$={e}^{-{x}^{2}}$

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

$\frac{dy}{dx}-2xy=1$

$\therefore {e}^{-{x}^{2}}\frac{dy}{dx}-2xy{e}^{-{x}^{2}}=1\cdot {e}^{-{x}^{2}}$

$\therefore \frac{d}{dx}\left(y{e}^{-{x}^{2}}\right)={e}^{-{x}^{2}}$

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

$y{e}^{-{x}^{2}}=\int {e}^{-{x}^{2}}dx$

The RHS integral does not have an elementary form, but we can use the definition of the Error Function :

$\text{erf}\left(x\right)=\frac{2}{\sqrt{\pi}}{\int}_{0}^{x}{e}^{-{t}^{2}}dt$

Which gives us:

$y{e}^{-{x}^{2}}=\frac{\sqrt{\pi}}{2}\text{erf}\left(x\right)+A$

$y=\frac{\sqrt{\pi}}{2}{e}^{{x}^{2}}\text{erf}\left(x\right)+A{e}^{{x}^{2}}$

asked 2022-09-21

What is a solution to the differential equation $\frac{dy}{dx}=\sqrt{xy}\mathrm{sin}x$?

asked 2021-09-25

If $f\left(x\right)+{x}^{2}{\left[f\left(x\right)\right]}^{5}=34$ and f(1) = 2, find f '(1).

asked 2021-09-08

Find the directional derivative of f at the given point in the direction indicated by the angle $\theta .f(x,y)={e}^{x}\mathrm{cos}y,(0,0),\theta =\frac{\pi}{4}$

asked 2022-07-01

I am trying to solve a first order differential equation with the condition that $g(y)=0$ if $y=0$:

$\begin{array}{rl}& a{g}^{\prime}(cy)+b{g}^{\prime}(ey)=\alpha \\ \text{(1)}& & g(0)=0,\end{array}$

where parameters a,b,c,e are real nonzero constants; $\alpha $ is a complex constant; function $g(y):\mathbb{R}\to \mathbb{C}$ is a function mapping from real number y to a complex number. The goal is to solve for function $g(\cdot )$. This is what I have done. Solve this differential equation by integrating with respect to y:

$\begin{array}{r}\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\alpha y+\beta ,\end{array}$

where $\beta $ is another complex constant. Plugging in y=0 and using the fact that g(0)=0, we have $\beta =0$. Therefore, we have

$\begin{array}{r}\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\alpha y.\end{array}$

The background of this problem is Cauchy functional equation, so my conjecture is one solution could be $g(y)=\gamma y$. Plugging in $g(y)=\gamma y$, I get $\gamma =\frac{\alpha}{a+b}$, which implies that one solution is $g(y)=\frac{\alpha}{a+b}y$. Then, I move on to show uniqueness. I define a vector-valued function $h\equiv ({h}_{1},{h}_{2}{)}^{T}$ such that

$\begin{array}{rl}{h}_{1}(y)& =\frac{a}{c}{g}_{1}(cy)+\frac{b}{e}{g}_{1}(ey)\\ {h}_{2}(y)& =\frac{a}{c}{g}_{2}(cy)+\frac{b}{e}{g}_{2}(ey),\end{array}$

where $g(y)\equiv {g}_{1}(y)+i{g}_{2}(y)$. Then, I rewrite this differential equation as

$\begin{array}{rl}& {h}^{\prime}(y)=\alpha \\ \text{(2)}& & h(0)=0,\end{array}$

where $\alpha \equiv ({\alpha}_{1},i{\alpha}_{2}{)}^{T}$. By the uniqueness theorem of first order differential equation, solution h(y) is unique. I have two questions. First, I think equation (1) and (2) should be equivalent. However, it seems that equation (1) can imply equation (2) but equation (2) may not imply equation (1). This is because h(0)=0 may imply either ${g}_{1}(0)=0,{g}_{2}(0)=0$ or $\frac{a}{c}+\frac{b}{e}=0$. Second, I have only proved that h(y) is unique. How should I proceed to show g(y) is also unique.

$\begin{array}{rl}& a{g}^{\prime}(cy)+b{g}^{\prime}(ey)=\alpha \\ \text{(1)}& & g(0)=0,\end{array}$

where parameters a,b,c,e are real nonzero constants; $\alpha $ is a complex constant; function $g(y):\mathbb{R}\to \mathbb{C}$ is a function mapping from real number y to a complex number. The goal is to solve for function $g(\cdot )$. This is what I have done. Solve this differential equation by integrating with respect to y:

$\begin{array}{r}\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\alpha y+\beta ,\end{array}$

where $\beta $ is another complex constant. Plugging in y=0 and using the fact that g(0)=0, we have $\beta =0$. Therefore, we have

$\begin{array}{r}\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\alpha y.\end{array}$

The background of this problem is Cauchy functional equation, so my conjecture is one solution could be $g(y)=\gamma y$. Plugging in $g(y)=\gamma y$, I get $\gamma =\frac{\alpha}{a+b}$, which implies that one solution is $g(y)=\frac{\alpha}{a+b}y$. Then, I move on to show uniqueness. I define a vector-valued function $h\equiv ({h}_{1},{h}_{2}{)}^{T}$ such that

$\begin{array}{rl}{h}_{1}(y)& =\frac{a}{c}{g}_{1}(cy)+\frac{b}{e}{g}_{1}(ey)\\ {h}_{2}(y)& =\frac{a}{c}{g}_{2}(cy)+\frac{b}{e}{g}_{2}(ey),\end{array}$

where $g(y)\equiv {g}_{1}(y)+i{g}_{2}(y)$. Then, I rewrite this differential equation as

$\begin{array}{rl}& {h}^{\prime}(y)=\alpha \\ \text{(2)}& & h(0)=0,\end{array}$

where $\alpha \equiv ({\alpha}_{1},i{\alpha}_{2}{)}^{T}$. By the uniqueness theorem of first order differential equation, solution h(y) is unique. I have two questions. First, I think equation (1) and (2) should be equivalent. However, it seems that equation (1) can imply equation (2) but equation (2) may not imply equation (1). This is because h(0)=0 may imply either ${g}_{1}(0)=0,{g}_{2}(0)=0$ or $\frac{a}{c}+\frac{b}{e}=0$. Second, I have only proved that h(y) is unique. How should I proceed to show g(y) is also unique.

asked 2022-09-11

Solve the differential equation $2x\mathrm{ln}x\frac{dy}{dx}+y=0$?

asked 2022-02-17

I am trying to solve a differential equation of the type $x{}^{\u2033}=-x+{x}^{3}$ . Now when I first integrate it with respect to time, $t$ , then I am getting ${x}^{\prime}=x({x}^{2}-1)+C$ , which is a non-linear first order differential equation. Now if the constant happens to be zero then I can solve it by partial fraction method but as life is not easier that constant term tagged to the $x}^{\prime$ is actually not zero. So how to proceed forward in this case.

Thank you!

Thank you!

asked 2022-09-04

How to you find the general solution of $\frac{dy}{dx}=\frac{{e}^{x}}{1+{e}^{x}}$?