$(1-{t}^{2}){y}^{t}-4ty=1,y(2)=-1$

What is the largest interval on which its solution is guaranteed to uniquely exist?

Francisco Proctor
2022-07-25
Answered

Solve the following initial value problem:

$(1-{t}^{2}){y}^{t}-4ty=1,y(2)=-1$

What is the largest interval on which its solution is guaranteed to uniquely exist?

$(1-{t}^{2}){y}^{t}-4ty=1,y(2)=-1$

What is the largest interval on which its solution is guaranteed to uniquely exist?

You can still ask an expert for help

neobuzdanio

Answered 2022-07-26
Author has **13** answers

The homogeneous differential equation can be solved by separation of variables.

$dy/y=4tdt/(1-{t}^{2})$ is the separated differential equation.

Integration $\mathrm{ln}(y)=-2\mathrm{ln}(-1+{t}^{2})$. Exponentiation: $y(t)=a({t}^{2}-1{)}^{-2}.y(2)=a/9=-1.a=-9.y(t)=-9/({t}^{2}-1{)}^{2}$.

The inhomogeneous differential equation is solved by:$y(t)=(-25+3t-{t}^{2})/(3(-1+{t}^{2}{)}^{2})$.This is derived by variation of constants.$y(t)=f(t)(-1+{t}^{2}{)}^{-2}.{y}^{\prime}(t)={f}^{\prime}(t)(-1+{t}^{2}{)}^{2}-4tf(t)(-1+{t}^{2}{)}^{-3}$.So in the inhomogeneous differential equation: $-{f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}+4tf(t)(-1+{t}^{2}{)}^{-2}-4tf(t)(-1+{t}^{2}{)}^{-2}={f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}=1$. Simply integrate and adopt the constant to rewrite the above result.

$dy/y=4tdt/(1-{t}^{2})$ is the separated differential equation.

Integration $\mathrm{ln}(y)=-2\mathrm{ln}(-1+{t}^{2})$. Exponentiation: $y(t)=a({t}^{2}-1{)}^{-2}.y(2)=a/9=-1.a=-9.y(t)=-9/({t}^{2}-1{)}^{2}$.

The inhomogeneous differential equation is solved by:$y(t)=(-25+3t-{t}^{2})/(3(-1+{t}^{2}{)}^{2})$.This is derived by variation of constants.$y(t)=f(t)(-1+{t}^{2}{)}^{-2}.{y}^{\prime}(t)={f}^{\prime}(t)(-1+{t}^{2}{)}^{2}-4tf(t)(-1+{t}^{2}{)}^{-3}$.So in the inhomogeneous differential equation: $-{f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}+4tf(t)(-1+{t}^{2}{)}^{-2}-4tf(t)(-1+{t}^{2}{)}^{-2}={f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}=1$. Simply integrate and adopt the constant to rewrite the above result.

asked 2021-01-02

Find

asked 2021-11-25

Represent the plane curve by a vector-valued function.

$2x-3y+5=0$

asked 2020-12-16

Find the gradient at the point (3,-3,2) of the scaler field given by

asked 2022-04-27

How to solve the ODE

$y\text{'}=\frac{x-{e}^{x}}{x+{e}^{y}}$

asked 2022-07-23

The temperature $T(x)$ at each point $x$ on the surface of Mars (a sphere) is a continuous function. Show that there is a point $x$ on the surface such that $T(x)=T(-x)$

(Hint: Represent the surface of Mars as $\{x\in {\mathbb{R}}^{3}:||x||=1\}$.)

Consider the function $f(x)=T(x)-T(-x)$

So.....

I consider an unit sphere is locating at the origin of a $xyz$-plane.

As $||x||=1$, I can say with $radius=1=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$

To find there is a point $T(x)=T(-x)$, we use the formula $f(x)=T(x)-T(-x)$ and show somehow $f(x)$ will equal to 0 ???

It will be a point in the upper hemisphere and another point with the exactly opposite vector (if using $ijk$ plane) on the lower hemisphere

(Hint: Represent the surface of Mars as $\{x\in {\mathbb{R}}^{3}:||x||=1\}$.)

Consider the function $f(x)=T(x)-T(-x)$

So.....

I consider an unit sphere is locating at the origin of a $xyz$-plane.

As $||x||=1$, I can say with $radius=1=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$

To find there is a point $T(x)=T(-x)$, we use the formula $f(x)=T(x)-T(-x)$ and show somehow $f(x)$ will equal to 0 ???

It will be a point in the upper hemisphere and another point with the exactly opposite vector (if using $ijk$ plane) on the lower hemisphere

asked 2021-11-27

Find two different sets of parametric equations for a rectangular equation

$y-{x}^{2}-3$

asked 2022-07-14

Rate of change of cross-section of cylinder

I got this task on my calculus class and I got stuck at process of figuring it out

A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.

What I know:

Volume is unknown and constant

Rate of change of diameter is

$\frac{dD}{dt}}=0,2\text{cm}/\text{sec$

Rate of change of height is

$\frac{dh}{dt}}=-0,4\text{cm}/\text{sec$

Trying to find $\frac{dD}{dt}$ by using formula

$V=\pi {r}^{2}h$

I got this task on my calculus class and I got stuck at process of figuring it out

A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.

What I know:

Volume is unknown and constant

Rate of change of diameter is

$\frac{dD}{dt}}=0,2\text{cm}/\text{sec$

Rate of change of height is

$\frac{dh}{dt}}=-0,4\text{cm}/\text{sec$

Trying to find $\frac{dD}{dt}$ by using formula

$V=\pi {r}^{2}h$