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kobusy [5.1K]
3 years ago
13

Subject differential equationDay. Month.Year. ..(y-secx) dx + tanxdy...o​

Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0

(y-\sec x)\,\mathrm dx+\tan x\,\mathrm dy=0

Divide both side by \mathrm dx and rearrange terms to get a linear ODE;

\tan x\dfrac{\mathrm dy}{\mathrm dx}+y=\sec x

Multiply both sides by \cos x:

\sin x\dfrac{\mathrm dy}{\mathrm dx}+\cos x\,y=1

The left side can be condensed as the derivative of a product:

\dfrac{\mathrm d}{\mathrm dx}(\sin x\,y)=1

Integrate both sides, then solve for y:

\sin x\,y=x+C\implies\boxed{y(x)=x\csc x+C\csc x}

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What is the slope of the line passing through the points (1, 2) and (5, 4)?​
bulgar [2K]
<h2>SOLVING</h2>

\Large\maltese\underline{\textsf{A. What is Asked}}

What is the slope of the line passing through the point (1,2) and (5,4)

\Large\maltese\underline{\textsf{This problem has been solved!}}

Formula used, here  \bf{\dfrac{y2-y1}{x2-x1}

_______________________________________________________

\bf{\dfrac{4-2}{5-1} | simplify

\bf{\dfrac{2}{4} | reduce

\bf{\dfrac{1}{2}

\rule{300}{1.7}

\bf{Result:}

         \bf{=Slope:\dfrac{1}{2}

\boxed{\bf{aesthetic\not101}}

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Step-by-step explanation:

3 0
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The value of x and y of given equation by using cramer's rule; 2x-y=5, x-2y=1​
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Step-by-step explanation:

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ᵃ ᵇ

ᶜ ᵈ

5 0
4 years ago
find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration
Natasha_Volkova [10]

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]

now for given values:

\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\

\to  (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to  (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} -  \frac{1}{4}] =0 \\\\\to  (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\

\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\

       = max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}

In point b:

Its  

spectral radius is less than 1 hence matrix is convergent.

In point c:

\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) =   \left(\begin{array}{c}3&1&2\end{array}\right)  , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\  \to x^{(k+1)} =  \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right]  \\\\

after solving the value the answer is

:

\lim_{k \to \infty} x^k=o  = \left[\begin{array}{c}0&0&0\end{array}\right]

4 0
3 years ago
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