 Recent questions in Differential equations
2022-01-19

A bacterial population B is known to have a rate of growth proportional to B itself. If between noon and 2pm the population triples, at what time no controls being exerted, should B becomes 100 times what it was at noon? what wiil be the population at 6pm? berljivx8 2022-01-18 Answered

Finding eigenvalues by inspection? I need to solve the following problem, In this problem, the eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system. $$\displaystyle{{x}_{{{1}}}^{{'}}}={2}{x}_{{{1}}}+{x}_{{{2}}}-{x}_{{{3}}}$$ $$\displaystyle{{x}_{{{2}}}^{{'}}}=-{4}{x}_{{{1}}}-{3}{x}_{{{2}}}-{x}_{{{3}}}$$ $$\displaystyle{{x}_{{{3}}}^{{'}}}={4}{x}_{{{1}}}+{4}{x}_{{{2}}}+{2}{x}_{{{3}}}$$ Now I know how to find the eigenvalues by using the fact that $$\displaystyle{\left|{A}-\lambda{I}\right|}={0}$$, but how would I do it by inspection? Inspection is easy for matrices that have the sum of their rows adding up to the same value, but this coefficient matrix doesn't have that property. EDIT: Originally I didn't understand what inspection meant either. After googling it this is what I found. Imagine you have the matrix, $A=(\begin{array}{c}2&-1&-1\\ -1&2&-1\\-1&-1&2\end{array})$ By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0. ajedrezlaproa6j 2022-01-18 Answered

Is there a closed form for any function f(x,y) satisfying: $$\displaystyle{\frac{{{d}{f}}}{{{\left.{d}{x}\right.}}}}+{\frac{{{d}{f}}}{{{\left.{d}{y}\right.}}}}={x}{y}$$ Linda Seales 2022-01-18 Answered

Obtaining Differential Equations from Functions $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}^{{{2}}}-{1}$$ is a first order ODE, $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}+{2}{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}^{{{2}}}+{y}={0}$$ is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for $$\displaystyle{y}={e}^{{{x}}}{\left({A}{\cos{{x}}}+{B}{\sin{{x}}}\right)}$$ and the steps that I followed are as follows. $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={e}^{{{x}}}{\left({A}{\cos{{x}}}+{B}{\sin{{x}}}\right)}+{e}^{{{x}}}{\left(-{A}{\sin{{x}}}+{B}{\cos{{x}}}\right)}$$ $$\displaystyle={y}+{e}^{{{x}}}{\left(-{A}{\sin{{x}}}+{B}{\cos{{x}}}\right)}$$ (1) $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}{\left(-{A}{\sin{{x}}}+{B}{\cos{{x}}}\right)}+{e}^{{{x}}}{\left(-{A}{\cos{{x}}}-{B}{\sin{{x}}}\right)}$$ $$\displaystyle={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{y}\right)}-{y}$$ using the orginal function and (1). Finally, $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}-{2}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{2}{y}={0}$$, which is the required differential equation. Similarly, if the function is $$y=(A \cos 2t+B \sin 2t)$$, the differential equation that I get is $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}+{4}{y}={0}$$ following similar steps as above. Patricia Crane 2022-01-18 Answered

Treacherous Euler-Lagrange equation $$\displaystyle{\left({y}'\right)}^{{{2}}}={2}{\left({1}-{\cos{{\left({y}\right)}}}\right)}$$ where y is a function of x subjected to boundary conditions $$\displaystyle{y}{\left({x}\right)}\rightarrow{0}\ \text{as}\ {x}\rightarrow-\infty\ \text{and}\ {y}{\left({x}\right)}\rightarrow{2}\pi\ \text{as}\ {x}\rightarrow+\infty$$, how might I find all its solutions? rheisf 2022-01-18 Answered

Reverse Laplace transform of $$\displaystyle{\frac{{{3}{s}-{15}}}{{{2}{s}^{{{2}}}-{4}{s}+{10}}}}$$ veksetz 2022-01-18 Answered

I have the following equation $$\displaystyle{\left({x}{y}^{{{2}}}+{x}\right)}{\left.{d}{x}\right.}+{\left({y}{x}^{{{2}}}+{y}\right)}{\left.{d}{y}\right.}={0}$$ and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method. Let $$\displaystyle{M}={x}{y}^{{{2}}}+{x}\ \text{and}\ {N}={y}{x}^{{{2}}}+{y}$$ $$\displaystyle{M}{y}={2}{x}{y}\ \text{and}\ {N}{x}={2}{x}{y}$$ $$\displaystyle\int{M}.{\left.{d}{x}\right.}\Rightarrow\int{x}{y}^{{{2}}}+{x}={x}^{{{2}}}{y}^{{{2}}}+\frac{{{x}^{{{2}}}}}{{2}}+{g{{\left({y}\right)}}}$$ Partial of $$\displaystyle{\left({x}^{{{2}}}{y}^{{{2}}}+\frac{{{x}^{{{2}}}}}{{2}}+{g{{\left({y}\right)}}}\right)}\Rightarrow{x}{y}^{{{2}}}+{g{{\left({y}\right)}}}'$$ $$\displaystyle{g{{\left({y}\right)}}}'={y}$$ $$\displaystyle{g{{\left({y}\right)}}}=\frac{{y}^{{{2}}}}{{2}}$$ the general solution then is $$\displaystyle{C}={x}^{{{2}}}\frac{{y}^{{{2}}}}{{2}}+\frac{{x}^{{{2}}}}{{2}}+\frac{{y}^{{{2}}}}{{2}}$$ Is this solution the same I would get if I had taken the Separate Equations route? Agohofidov6 2022-01-18 Answered

I'm using the method of undetermined coefficients to find a particular solution of: $$\displaystyle{y}{''}+{y}'={x}{e}^{{-{x}}}$$ Ostensibly, it seems that $$\displaystyle{y}_{{{p}}}$$ should take the form of $$\displaystyle{\left({A}{x}+{B}\right)}{e}^{{-{x}}}$$ At least that's the form that I think I've been taught. Problem is that it just doesn't work out for me. I get a value for A, but not for B... Am I choosing an incorrect yp form? James Dale 2022-01-18 Answered

I want to get a particular solution to the differential equation $$\displaystyle{y}{''}+{2}{y}'+{2}{y}={2}{e}^{{{x}}}{\cos{{\left({x}\right)}}}$$ and therefore I would like to 'complexify' the right hand side. This means that I want to write the right hand side as $$\displaystyle{q}{\left({x}\right)}{e}^{{\alpha{x}}}$$ with q(x) a polynomial. How is this possible? jubateee 2022-01-18 Answered

Differential equation $$\displaystyle{f}{''}{\left({x}\right)}+{\frac{{{\left({n}-{1}\right)}{\left({f}'{\left({x}\right)}\right)}^{{{2}}}}}{{{\sin{{h}}}{\left({x}\right)}}}}={0}$$ zakinutuzi 2022-01-18 Answered

$$\displaystyle{y}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={e}^{{{x}}}$$ Solve it by ''separating variables'' like this: $$\displaystyle{y}{\left.{d}{y}\right.}={e}^{{{x}}}{\left.{d}{x}\right.}$$ $$\displaystyle\int{y}{\left.{d}{y}\right.}=\int{e}^{{{x}}}{\left.{d}{x}\right.}$$ $$\displaystyle\frac{{y}^{{{2}}}}{{2}}={e}^{{{x}}}+{c}$$ Kathleen Rausch 2022-01-18 Answered

How would I solve these differential equations? Thanks so much for the help! $$\displaystyle{{P}_{{{0}}}^{{'}}}{\left({t}\right)}=\alpha{P}_{{{1}}}{\left({t}\right)}-\beta{P}_{{{0}}}{\left({t}\right)}$$ $$\displaystyle{{P}_{{{1}}}^{{'}}}=\beta{P}_{{{0}}}{\left({t}\right)}-\alpha{P}_{{{1}}}{\left({t}\right)}$$ We also know $$\displaystyle{P}_{{{0}}}{\left({t}\right)}+{P}_{{{1}}}{\left({t}\right)}={1}$$ Joanna Benson 2022-01-18 Answered

Solve: $$\displaystyle{y}{''}+{2}{y}'-{3}{y}={0}$$ Mabel Breault 2022-01-18 Answered

Nonlinear Ordinary differential equation. $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}-{\frac{{{2}}}{{{y}^{{{2}}}}}}={0}$$ with $$\displaystyle{y}{\left({0}\right)}={a}\ \text{and}\ {y}'{\left({0}\right)}={0}$$ Where a is a known constant. maduregimc 2022-01-18 Answered

Applications of Differential and Difference Equations Solve $$\displaystyle{\left({D}^{{{2}}}-{2}{D}+{1}\right)}{y}={x}{e}^{{{x}}}{\sin{{x}}}$$ Shirley Thompson 2022-01-18 Answered

Which of the following differential equations has two real and equal roots? Select one: a. $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={4}{y}$$ b. $$\displaystyle{16}{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}-{8}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{y}={0}$$ c. $$\displaystyle{3}{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}+{2}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}$$ d. $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}=-{4}{y}$$ deiteresfp 2022-01-17 Answered

Solve the following equation? $$\displaystyle{\left({x}-{2}{x}{y}-{y}^{{{2}}}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{y}^{{{2}}}={0}$$ zgribestika 2022-01-17 Answered

How do you solve the Initial value probelm $$\displaystyle{\frac{{{d}{p}}}{{{\left.{d}{t}\right.}}}}={10}{p}{\left({1}-{p}\right)}$$, $$\displaystyle{p}{\left({0}\right)}={0.1}$$ Solve and show that $$\displaystyle{p}{\left({t}\right)}\rightarrow{1}\ \text{as}\ {t}\rightarrow\infty$$. Cheexorgeny 2022-01-17 Answered

Solving second-order nonlinear ordinary differential equation $$\displaystyle{y}{''}{\left({x}\right)}={\frac{{{4}}}{{{3}}}}{y}{\left({x}\right)}^{{{3}}}{y}'{\left({x}\right)}$$ given that $$\displaystyle{y}{\left({0}\right)}={1}\ \text{and}\ {y}'{\left({0}\right)}={\frac{{{1}}}{{{3}}}}$$. Agohofidov6 2022-01-17 Answered

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