prsategazd
2022-01-20
Answered

Solve the first-order differential equation
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${x}^{2}{y}^{2}dx-({x}^{3}+1)dy=0$

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asked 2022-01-21

The equation ${y}^{\prime}=\frac{y}{x}(\mathrm{ln}y-\mathrm{ln}x+1)$ is?

1. First order, Partial.

2. First order, Non-homogenous.

3. None of above.

4. First order, Homogenous.

1. First order, Partial.

2. First order, Non-homogenous.

3. None of above.

4. First order, Homogenous.

asked 2021-09-18

Find the directional derivative of f at the given point in the direction indicated by the angle theta. $f(x,y)=y{e}^{-x},(0,4),\theta =\frac{2\pi}{3}$

asked 2022-09-11

Solve $\frac{dy}{dx}-\frac{1}{2}(1+\frac{1}{x})y+\frac{3}{x}{y}^{3}=0$?

asked 2022-05-27

If I have a differential equation ${y}^{\prime}(t)=Ay(t)$ where A is a constant square matrix that is not diagonalizable(although it is surely possible to calculate the eigenvalues) and no initial condition is given. And now I am interested in the fundamental matrix. Is there a general method to determine this matrix? I do not want to use the exponential function and the Jordan normal form, as this is quite exhausting. Maybe there is also an ansatz possible as it is for the special case, where this differential equation is equivalent to an n-th order ode. I saw a method where they calculated the eigenvalues of the matrix and depending on the multiplicity n of this eigenvalue they used an exponential term(with the eigenvalue) and in each component an n-th order polynomial as a possible ansatz. Though they only did this, when they were interested in a initial value problem, so with an initial condition and not for a general solution.

I was asked to deliver an example: so ${y}^{\prime}(t)=\left(\begin{array}{cc}3& -4\\ 1& -1\end{array}\right)y(t)$ If somebody can construct a fundamental matrix for this system, than this should be sufficient

I was asked to deliver an example: so ${y}^{\prime}(t)=\left(\begin{array}{cc}3& -4\\ 1& -1\end{array}\right)y(t)$ If somebody can construct a fundamental matrix for this system, than this should be sufficient

asked 2022-06-29

I have been trying to solve a differential equation as a practice question for my test, but I am just unable to get the correct answer. Please have a look at the D.E:

$dy/dx=1/(3x+\mathrm{sin}(3y))$

My working is as follows:

$dx/dy=3x+\mathrm{sin}(3y)$

$dx-3x=\mathrm{sin}(3y)dy$

Integrating both sides:

$x-(3/2){x}^{2}=-(1/3)\mathrm{cos}(3y)+c$

But the correct answer is:

$x=c{e}^{3}y-1/6(\mathrm{cos}(3y)+\mathrm{sin}(3y))$, which is quite different from what I have got. Could someone please help me solve this?

$dy/dx=1/(3x+\mathrm{sin}(3y))$

My working is as follows:

$dx/dy=3x+\mathrm{sin}(3y)$

$dx-3x=\mathrm{sin}(3y)dy$

Integrating both sides:

$x-(3/2){x}^{2}=-(1/3)\mathrm{cos}(3y)+c$

But the correct answer is:

$x=c{e}^{3}y-1/6(\mathrm{cos}(3y)+\mathrm{sin}(3y))$, which is quite different from what I have got. Could someone please help me solve this?

asked 2022-07-02

I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in Matlab

${y}^{\prime}-xy=x{y}^{3/2}\phantom{\rule{thinmathspace}{0ex}},y(1)=4$

$\begin{array}{rlr}{y}^{\prime}-xy=& x{y}^{3/2}& \\ {\displaystyle \frac{dy}{dx}}=& x(y+{y}^{3/2})& \\ {\displaystyle \frac{dy}{(y+{y}^{3/2})}}=& x\phantom{\rule{thinmathspace}{0ex}}dx& \\ -2\mathrm{ln}{\displaystyle \frac{1+\sqrt{y}}{\sqrt{y}}}=& {\displaystyle \frac{{x}^{2}}{2}}+c& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}+c}{)}^{2}}}& \\ 4=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-1}{4}}+c}{)}^{2}}}& \\ c=& -0.655& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}-0.655}{)}^{2}}}& \end{array}$

This is what I'm getting in Matlab

$\left(\begin{array}{c}\frac{{(\text{tanh}(\frac{{x}^{2}}{8}+\text{atanh}\left(3\right)-\frac{1}{8})+1)}^{2}}{4}\\ \frac{{(\text{tanh}(-\frac{{x}^{2}}{8}+\text{atanh}\left(5\right)+\frac{1}{8})-1)}^{2}}{4}\end{array}\right)$

${y}^{\prime}-xy=x{y}^{3/2}\phantom{\rule{thinmathspace}{0ex}},y(1)=4$

$\begin{array}{rlr}{y}^{\prime}-xy=& x{y}^{3/2}& \\ {\displaystyle \frac{dy}{dx}}=& x(y+{y}^{3/2})& \\ {\displaystyle \frac{dy}{(y+{y}^{3/2})}}=& x\phantom{\rule{thinmathspace}{0ex}}dx& \\ -2\mathrm{ln}{\displaystyle \frac{1+\sqrt{y}}{\sqrt{y}}}=& {\displaystyle \frac{{x}^{2}}{2}}+c& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}+c}{)}^{2}}}& \\ 4=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-1}{4}}+c}{)}^{2}}}& \\ c=& -0.655& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}-0.655}{)}^{2}}}& \end{array}$

This is what I'm getting in Matlab

$\left(\begin{array}{c}\frac{{(\text{tanh}(\frac{{x}^{2}}{8}+\text{atanh}\left(3\right)-\frac{1}{8})+1)}^{2}}{4}\\ \frac{{(\text{tanh}(-\frac{{x}^{2}}{8}+\text{atanh}\left(5\right)+\frac{1}{8})-1)}^{2}}{4}\end{array}\right)$

asked 2022-09-12

$xy\prime -3y=x-1$ Solve