Janet Hart
2022-09-26
Answered

How do you solve the differential $\frac{dy}{dx}=\frac{x-4}{\sqrt{{x}^{2}-8x+1}}$?

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Lilia Horton

Answered 2022-09-27
Author has **6** answers

$\frac{dy}{dx}=\frac{x-4}{\sqrt{{x}^{2}-8x+1}}$

Is a First Order separable DE which we can sole by integrating:

$\therefore y=\int \frac{x-4}{\sqrt{{x}^{2}-8x+1}}dx$

Let $u={x}^{2}-8x+1\Rightarrow \frac{du}{dx}=2x-8=2(x-4)$

Substituting into the RHS integral we get:

$y=\int \frac{\frac{1}{2}}{\sqrt{u}}du$

$=\frac{1}{2}\int {u}^{-\frac{1}{2}}du$

$=\frac{1}{2}\frac{{u}^{\frac{1}{2}}}{\frac{1}{2}}+C$

$=\sqrt{u}+C$

$=\sqrt{{x}^{2}-8x+1}+C$

which is the General Solution

Is a First Order separable DE which we can sole by integrating:

$\therefore y=\int \frac{x-4}{\sqrt{{x}^{2}-8x+1}}dx$

Let $u={x}^{2}-8x+1\Rightarrow \frac{du}{dx}=2x-8=2(x-4)$

Substituting into the RHS integral we get:

$y=\int \frac{\frac{1}{2}}{\sqrt{u}}du$

$=\frac{1}{2}\int {u}^{-\frac{1}{2}}du$

$=\frac{1}{2}\frac{{u}^{\frac{1}{2}}}{\frac{1}{2}}+C$

$=\sqrt{u}+C$

$=\sqrt{{x}^{2}-8x+1}+C$

which is the General Solution

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My Problem is this given System of differential Equations:

$\dot{x}=8x+18y$

$\dot{y}=-3x-7y$

I am looking for a gerenal solution.

My Approach was: i can see this is a System of linear and ordinary differential equations. Both are of first-order, because the highest derivative is the first. But now i am stuck, i have no idea how to solve it. A Transformation into a Matrix should lead to this expression:

$\overrightarrow{y}=\left(\begin{array}{cc}8& 18\\ -3& -7\end{array}\right)\cdot x$

or is this correct:

$\overrightarrow{x}=\left(\begin{array}{cc}8& 18\\ -3& -7\end{array}\right)\cdot y\text{?}$

But i don't know how to determine the solution, from this point on.

$\dot{x}=8x+18y$

$\dot{y}=-3x-7y$

I am looking for a gerenal solution.

My Approach was: i can see this is a System of linear and ordinary differential equations. Both are of first-order, because the highest derivative is the first. But now i am stuck, i have no idea how to solve it. A Transformation into a Matrix should lead to this expression:

$\overrightarrow{y}=\left(\begin{array}{cc}8& 18\\ -3& -7\end{array}\right)\cdot x$

or is this correct:

$\overrightarrow{x}=\left(\begin{array}{cc}8& 18\\ -3& -7\end{array}\right)\cdot y\text{?}$

But i don't know how to determine the solution, from this point on.

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A large tank is partially filled with 100 gallons of fluid in which 10 pounds of

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the tank at a rate of 6 gal/min. The well-mixed solution is then pumped out

at a slower rate of 4 gal/min. Find the number of pounds of salt in the tank

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Let

$\left[\begin{array}{ccc}-2& 1& 0\\ 0& -2& 1\\ 0& 0& -2\end{array}\right]$

and

$|x(t)|=({x}_{1}^{2}(t)+{x}_{2}^{2}(t)+{x}_{3}^{2}(t){)}^{1/2}$

Then any solution of the first order system of the ordinary differential equation

$\{\begin{array}{r}{x}^{\prime}(t)=Ax(t)\\ x(0)={x}_{0}\end{array}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}$

satisfies

1. $\underset{t\to \mathrm{\infty}|x(t)|=0}{lim}$

2. $\underset{t\to \mathrm{\infty}|x(t)|=\mathrm{\infty}}{lim}$

3. $\underset{t\to \mathrm{\infty}|x(t)|=2}{lim}$

4. $\underset{t\to \mathrm{\infty}|x(t)|=12}{lim}$

$\left[\begin{array}{ccc}-2& 1& 0\\ 0& -2& 1\\ 0& 0& -2\end{array}\right]$

and

$|x(t)|=({x}_{1}^{2}(t)+{x}_{2}^{2}(t)+{x}_{3}^{2}(t){)}^{1/2}$

Then any solution of the first order system of the ordinary differential equation

$\{\begin{array}{r}{x}^{\prime}(t)=Ax(t)\\ x(0)={x}_{0}\end{array}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}$

satisfies

1. $\underset{t\to \mathrm{\infty}|x(t)|=0}{lim}$

2. $\underset{t\to \mathrm{\infty}|x(t)|=\mathrm{\infty}}{lim}$

3. $\underset{t\to \mathrm{\infty}|x(t)|=2}{lim}$

4. $\underset{t\to \mathrm{\infty}|x(t)|=12}{lim}$

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