Inyalan0
2021-12-30
Answered

Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification. $2x{\mathrm{sin}}^{2}ydx-({x}^{2}-9)\mathrm{cos}ydy=0$

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Philip Williams

Answered 2021-12-31
Author has **39** answers

Given differential equation

$2x{\mathrm{sin}}^{2}ydx-({x}^{2}-9)\mathrm{cos}ydy=0$

write the equation in the standard form

$\frac{dy}{dx}=\frac{2x{\mathrm{sin}}^{2}y}{({x}^{2}-9)\mathrm{cos}y}$

we can saperate the variable so, equation is seperable now seperating the variable

$\left(\frac{\mathrm{cos}y}{{\mathrm{sin}}^{2}y}\right)dy=\left(\frac{2x}{{x}^{2}-9}\right)dx$

integrate both side

$\Rightarrow \int \left(\frac{\mathrm{cos}y}{{\mathrm{sin}}^{2}y}\right)dy=\int \left(\frac{2x}{{x}^{2}-9}\right)dx$

$\Rightarrow \int \frac{d\left(\mathrm{sin}y\right)}{{\mathrm{sin}}^{2}y}=\int \frac{d({x}^{2}-9)}{{x}^{2}-9}$

$\Rightarrow -\frac{1}{\mathrm{sin}y}=\mathrm{ln}({x}^{2}-9)+C$

$\Rightarrow \mathrm{ln}({x}^{2}-9)+\mathrm{cos}\text{}ec\text{}y=C$

hence solution of differential equation is

$\mathrm{ln}({x}^{2}-9)+\mathrm{cos}\text{}ec\text{}y=C$

write the equation in the standard form

we can saperate the variable so, equation is seperable now seperating the variable

integrate both side

hence solution of differential equation is

temnimam2

Answered 2022-01-01
Author has **36** answers

This is variable separable form.

This is general solution.

karton

Answered 2022-01-09
Author has **439** answers

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$\begin{array}{}\text{(1)}& A\frac{\mathrm{\partial}}{\mathrm{\partial}t}g(t)+Bg(t)=f(t),\end{array}$

where A and B are constants, g(t) is associated with the reflected wave, and f(t) is a (finite) driving function associated with the incident wave. Both A and B may be positive or negative. I am interested in the behavior of the solution in the limit that $A\to 0$

In studying a reflection-transmission problem involving exotic materials, I have come across the following linear first-order differential equation:A∂∂tg(t)+Bg(t)=f(t),(1)where A and B are constants, g(t) is associated with the reflected wave, and f(t) is a (finite) driving function associated with the incident wave. Both A and B may be positive or negative. I am interested in the behavior of the solution in the limit that A\rightarrow0.

I know there is an exact solution to Eq. (1), which is

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where C=0 because g(t)=0 if f(t)=0. However, I do not understand how this exact solution reduces to the case where A=0, which is $g(t)={B}^{-1}f(t)$. Any insight would be greatly appreciated.

I've seen a lot of documents discussing asymptotic analyses of linear differential equations, but they all start with second-order equations. Is this because there is inherently problematic with first-order?

$\begin{array}{}\text{(1)}& A\frac{\mathrm{\partial}}{\mathrm{\partial}t}g(t)+Bg(t)=f(t),\end{array}$

where A and B are constants, g(t) is associated with the reflected wave, and f(t) is a (finite) driving function associated with the incident wave. Both A and B may be positive or negative. I am interested in the behavior of the solution in the limit that $A\to 0$

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I know there is an exact solution to Eq. (1), which is

$g(t)=C{e}^{-Bt/A}+\frac{1}{A}{\int}_{-\mathrm{\infty}}^{t}{e}^{-B(t-{t}^{\prime})/A}f({t}^{\prime})d{t}^{\prime},$

where C=0 because g(t)=0 if f(t)=0. However, I do not understand how this exact solution reduces to the case where A=0, which is $g(t)={B}^{-1}f(t)$. Any insight would be greatly appreciated.

I've seen a lot of documents discussing asymptotic analyses of linear differential equations, but they all start with second-order equations. Is this because there is inherently problematic with first-order?

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