Solve in two different methods.

$(y-x+3)dx+(x+y+3)dy=0;\text{}x=1;\text{}y=4.$

veksetz
2022-01-22
Answered

Solve in two different methods.

$(y-x+3)dx+(x+y+3)dy=0;\text{}x=1;\text{}y=4.$

You can still ask an expert for help

Navreaiw

Answered 2022-01-22
Author has **34** answers

Its solution is

x=1, y=4

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Find the general solution of the given second-order differential equations

$y{}^{\u2033}+4{y}^{\prime}-y=0$

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Solve by using the Laplace Transform Method?

${y}^{\prime}+y=6{e}^{{t}^{2}},\text{}\text{}\text{}y\left(0\right)=3$

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Find the inverse Laplace transform of

$\frac{6s+15}{({s}^{2}+25)}$

$s>0$

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A simple question about the solution of homogeneous equation to this differential equation

Given that$t,1+t,{t}^{2},-t$ are the solutions to $y{}^{\u2034}+a\left(t\right)y{}^{\u2033}+b\left(t\right)y{}^{\u2033}+c\left(t\right)y=d\left(t\right)$ , what is the solution of homogeneous equation to this differential equation? What i have done is tried the properties of linear differential equation that

$L\left(t\right)=L(1+t)=L\left({t}^{2}\right)=L(-t)=d\left(t\right)$ so the homogeneous solution should be independent and i claim that $1,t,{t}^{2}$ should be the solution. However, i am not sure hot can i actually conclude that these are the solutions? It seems that it can be quite a number of sets of solution by the linearity.

Given that

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Solve the initial value problem for

r

as a vector function of t. Differential equation:

$\frac{dr}{dt}=-ti-tj-tk$

Initial condition:

r(0)=i+2j+3k

r

as a vector function of t. Differential equation:

Initial condition:

r(0)=i+2j+3k

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Solve the differential equation by variation of parameters

$y"+y=\mathrm{sin}x$

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Consider the function

$f\left(\mu \right)=\sum _{i=1}{({x}_{i}-\mu )}^{2},$ where

${x}_{i}=i,\text{}i=1,\text{}2,\text{}\cdots ,\text{}n$

What is the first and second derivative os$f\left(\mu \right)$ ?

What is the first and second derivative os