What is the general solution of the differential equation? : $\frac{{d}^{4}y}{{dx}^{4}}-4y=0$

Leonel Schwartz
2022-09-29
Answered

What is the general solution of the differential equation? : $\frac{{d}^{4}y}{{dx}^{4}}-4y=0$

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Bleha7s

Answered 2022-09-30
Author has **11** answers

With linear operator $D=\frac{d}{dx}$, this is:

$({D}^{4}-4)y=0$

$({D}^{2}-2)({D}^{2}+2)y=0$

$(D-\sqrt{2})(D+\sqrt{2})(D-i\sqrt{2})(D+i\sqrt{2})y=0$

Each of these factors has a simple solution, eg:

1. $(D-\sqrt{2})y=0\Rightarrow y\prime =\sqrt{2}y\Rightarrow y=\alpha {e}^{\sqrt{2}x}$

2. $(D-i\sqrt{2})y=0\Rightarrow y\prime =i\sqrt{2}y\Rightarrow y=\beta {e}^{i\sqrt{2}x}$

The complete solution is the superposition of individual solutions:

$\Rightarrow y\left(x\right)=A{e}^{\sqrt{2}x}+B{e}^{-\sqrt{2}x}+C{e}^{i\sqrt{2}x}+D{e}^{-i\sqrt{2}x}$

$C{e}^{i\sqrt{2}x}+D{e}^{-i\sqrt{2}x}$ can also be re-written using the Euler formula , arriving at:

$y=A{e}^{\sqrt{2}x}+B{e}^{-\sqrt{2}x}+C\prime \mathrm{cos}\left(\sqrt{2}x\right)+D\prime \mathrm{sin}\left(\sqrt{2}x\right)$

$({D}^{4}-4)y=0$

$({D}^{2}-2)({D}^{2}+2)y=0$

$(D-\sqrt{2})(D+\sqrt{2})(D-i\sqrt{2})(D+i\sqrt{2})y=0$

Each of these factors has a simple solution, eg:

1. $(D-\sqrt{2})y=0\Rightarrow y\prime =\sqrt{2}y\Rightarrow y=\alpha {e}^{\sqrt{2}x}$

2. $(D-i\sqrt{2})y=0\Rightarrow y\prime =i\sqrt{2}y\Rightarrow y=\beta {e}^{i\sqrt{2}x}$

The complete solution is the superposition of individual solutions:

$\Rightarrow y\left(x\right)=A{e}^{\sqrt{2}x}+B{e}^{-\sqrt{2}x}+C{e}^{i\sqrt{2}x}+D{e}^{-i\sqrt{2}x}$

$C{e}^{i\sqrt{2}x}+D{e}^{-i\sqrt{2}x}$ can also be re-written using the Euler formula , arriving at:

$y=A{e}^{\sqrt{2}x}+B{e}^{-\sqrt{2}x}+C\prime \mathrm{cos}\left(\sqrt{2}x\right)+D\prime \mathrm{sin}\left(\sqrt{2}x\right)$

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How to solve the differential equation $(dy/dx{)}^{2}=(x-y{)}^{2}$ with initial condition $y(0)=0$?

I solved the equation by partitioning it into two differential equations.

1) $dy/dx=x-y$ The solution is $\to $ $1-x+y=-\mathrm{exp}(-x)$

and

2) $1+x-y=\mathrm{exp}(x)$

How do we write combined solution of such equations.

I solved the equation by partitioning it into two differential equations.

1) $dy/dx=x-y$ The solution is $\to $ $1-x+y=-\mathrm{exp}(-x)$

and

2) $1+x-y=\mathrm{exp}(x)$

How do we write combined solution of such equations.

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Below is a problem I did. However, it did not match the back of the book. I would like to know where I went wrong.

Problem:

Solve the following differential equation.

${y}^{\prime}=\frac{y-x}{x}$

Answer:

$\begin{array}{rl}\frac{dy}{dx}& ={\displaystyle \frac{y}{x}}-{\displaystyle \frac{x}{x}}={\displaystyle \frac{y}{x}}-1\\ y& =xv\\ {\displaystyle \frac{dy}{dx}}& =x{\displaystyle \frac{dv}{dx}}+v\\ x{\displaystyle \frac{dv}{dx}}+v& =v-1\\ x{\displaystyle \frac{dv}{dx}}& =-1\\ dv& =-{\displaystyle \frac{dx}{x}}\\ v& =-\mathrm{ln}x+c\\ {\displaystyle \frac{y}{x}}& =-\mathrm{ln}|x|+c\\ y& =-x\mathrm{ln}|x|+cx\end{array}$

The book's answer is:

$y=x\mathrm{ln}|\frac{k}{x}|$

Where did I go wrong?

Problem:

Solve the following differential equation.

${y}^{\prime}=\frac{y-x}{x}$

Answer:

$\begin{array}{rl}\frac{dy}{dx}& ={\displaystyle \frac{y}{x}}-{\displaystyle \frac{x}{x}}={\displaystyle \frac{y}{x}}-1\\ y& =xv\\ {\displaystyle \frac{dy}{dx}}& =x{\displaystyle \frac{dv}{dx}}+v\\ x{\displaystyle \frac{dv}{dx}}+v& =v-1\\ x{\displaystyle \frac{dv}{dx}}& =-1\\ dv& =-{\displaystyle \frac{dx}{x}}\\ v& =-\mathrm{ln}x+c\\ {\displaystyle \frac{y}{x}}& =-\mathrm{ln}|x|+c\\ y& =-x\mathrm{ln}|x|+cx\end{array}$

The book's answer is:

$y=x\mathrm{ln}|\frac{k}{x}|$

Where did I go wrong?

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What is a solution to the differential equation $\frac{dy}{dx}=\frac{x+y}{x-y}$?