How do I solve the nonlinear differential equation $y\prime =-\frac{x}{y}$ under initial condition y(1)=1?

Conrad Beltran
2022-09-20
Answered

How do I solve the nonlinear differential equation $y\prime =-\frac{x}{y}$ under initial condition y(1)=1?

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xjiaminhoxy4

Answered 2022-09-21
Author has **9** answers

This is a first order linear separable ordinary differential equation. If we collect terms we have:

$y\frac{dy}{dx}=-x$

We can now "separate the variables and integrate:

$\int ydy=\int -xdx$

$\therefore \frac{1}{2}{y}^{2}=-\frac{1}{2}{x}^{2}+C$

Using the initial condition y(1)=1 we find:

$\therefore \frac{1}{2}=-\frac{1}{2}+C=1\Rightarrow C=1$

Thus we have the particular solution:

$\frac{1}{2}{y}^{2}=-\frac{1}{2}{x}^{2}+1$

$\therefore {y}^{2}=-{x}^{2}+2$

Hence:

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

$y\frac{dy}{dx}=-x$

We can now "separate the variables and integrate:

$\int ydy=\int -xdx$

$\therefore \frac{1}{2}{y}^{2}=-\frac{1}{2}{x}^{2}+C$

Using the initial condition y(1)=1 we find:

$\therefore \frac{1}{2}=-\frac{1}{2}+C=1\Rightarrow C=1$

Thus we have the particular solution:

$\frac{1}{2}{y}^{2}=-\frac{1}{2}{x}^{2}+1$

$\therefore {y}^{2}=-{x}^{2}+2$

Hence:

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

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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.

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where $f(x,y)$ is a homogeneous function. I found some examples like $f(x,y)=(x+y{)}^{2}$ where it can be solved after converting it to Ricatti's equation.

thanks

$\frac{dy}{dx}}=f(x,y)$

where $f(x,y)$ is a homogeneous function. I found some examples like $f(x,y)=(x+y{)}^{2}$ where it can be solved after converting it to Ricatti's equation.

thanks

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