Donald Johnson
2021-12-15
Answered

How do you find $\frac{{d}^{2}y}{{dx}^{2}}$ by implicit differentiation where ${x}^{2}y+x{y}^{2}=3x$

You can still ask an expert for help

Stella Calderon

Answered 2021-12-16
Author has **35** answers

Differentiating

${x}^{2}y+x{y}^{2}=3x$

with respect to x we get

$2xy+{x}^{2}{y}^{\prime}+{y}^{2}+2xy{y}^{\prime}=3$

So we get

$y}^{\prime}=\frac{3-{y}^{2}-2xy}{{x}^{2}+2xy$

for the second derivative we obtain

$y{}^{\u2033}=\frac{(-2y{y}^{\prime}-2y-2x{y}^{\prime})(x+2xy)-(3-{y}^{2}-2xy)(2x+2y+2x{y}^{\prime})}{{({x}^{2}+2xy)}^{2}}$

Now plug the result for y' in this equation!

with respect to x we get

So we get

for the second derivative we obtain

Now plug the result for y' in this equation!

William Appel

Answered 2021-12-17
Author has **44** answers

Thanks for the answer.

asked 2021-01-02

Find

asked 2022-04-22

Value of differential equation at non existent points in the original function.

$y}^{2}+y={x}^{3$

asked 2022-04-03

Find a differential equation for $w\left(t\right)={\varphi}_{t}\left(z\right)=\frac{z+\mathrm{tan}t}{1-z\left(\mathrm{tan}t\right)}$

Usually the question is the other way around, but (D. H. Sattinger and O. L. Weaver, 1986 Lie Groups and Algebras with Applications to Physics, Geometry, and Mechanics) pose it like this.

Earlier we showed that$\varphi}_{t+s}={\varphi}_{t}\circ {\varphi}_{s$ .

Since tan is a periodic function, I thing that the differential equation they are looking for might be of order 2.

My efforts so far consists of calculating w'(t), and trying to identify w in there, but things get messy.

Usually the question is the other way around, but (D. H. Sattinger and O. L. Weaver, 1986 Lie Groups and Algebras with Applications to Physics, Geometry, and Mechanics) pose it like this.

Earlier we showed that

Since tan is a periodic function, I thing that the differential equation they are looking for might be of order 2.

My efforts so far consists of calculating w'(t), and trying to identify w in there, but things get messy.

asked 2022-04-23

How to solve a complicated ODE

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 2022-04-08

Finding solution of ODE using Laplace transform

with initial condition$y\left(0\right)=0$ .

${y}^{\prime}+2y=f\left(x\right)$ where $f\left(x\right)=0\text{}\text{if}\text{}x1\text{}\text{and}\text{}f\left(x\right)=1\text{}\text{if}\text{}0\le x\le 1$ .

I applied Laplace transform on both sides which gets

$sY\left(s\right)+2Y\left(s\right)=\frac{1-{e}^{-s}}{s}\Rightarrow Y\left(s\right)=\frac{1-{e}^{-s}}{s(s+2)}=\frac{1}{s(s+2)}-\frac{{e}^{-s}}{s(s+2)}$

Now I know inverse Laplace transform of$\frac{1}{s(s+2)}$ but how to find inverse Laplace transform of second term? Or can we solve above ODE using different method?

with initial condition

I applied Laplace transform on both sides which gets

Now I know inverse Laplace transform of

asked 2022-04-23

How many solutions does the ODE have?

Given:

$$\{\begin{array}{l}{y}^{\prime}-{a}^{2}({y}^{\prime}{)}^{3}-\frac{\mathrm{sin}(x)}{x+y}=0\\ y(0)=1\end{array}$$

Write how many solutions does the system have for $a=0$ and $a\ne 0$.

asked 2022-04-22

Function whose minimum is independent of the input?

I have a question that asks me to show that the functional

$I\left(x\right)={\int}_{\left\{{x}_{1}\right\}}^{{x}_{2}}({x}^{2}+3{y}^{2}){y}^{\prime}+2xydx$

has an extremal that is independent of the choice of y that joins two arbitrary points$({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$ .

So I tried solving for the extremal, which give me the euler lagrange equation:

$6y{y}^{\prime}+2x=2x+6y{y}^{\prime}$

That's not an error it seems that$\frac{d}{dx}\left(\frac{\partial F}{\partial {y}^{\prime}}\right)=\frac{\partial F}{\partial y}$ .

Ok then no wonder that the path doesn't matter. But now I am trying to compute the value of the functional for the two points, I tried a straight line between the 2, but this is giving me a messy expression, it's computable but ugly:

$J\left(y\right)={\int}_{\left\{{x}_{1}\right\}}^{{x}_{2}}s({x}^{2}+3{(sx+k)}^{2})+2x(sx+k)dx$

(where s and k are the linear coefficients of the line connecting the two points).

I have a question that asks me to show that the functional

has an extremal that is independent of the choice of y that joins two arbitrary points

So I tried solving for the extremal, which give me the euler lagrange equation:

That's not an error it seems that

Ok then no wonder that the path doesn't matter. But now I am trying to compute the value of the functional for the two points, I tried a straight line between the 2, but this is giving me a messy expression, it's computable but ugly:

(where s and k are the linear coefficients of the line connecting the two points).