If $y=x{e}^{7x}$ is a solution of a linear homogeneous 2nd order DE, then another solution might be?

capellitad9
2022-07-17
Answered

If $y=x{e}^{7x}$ is a solution of a linear homogeneous 2nd order DE, then another solution might be?

You can still ask an expert for help

Damarion Pierce

Answered 2022-07-18
Author has **11** answers

For a second order linear homogenous equations,

If roots of auxiliary equation are

1. real and distinct then the solution is

$y={C}_{1}{e}^{{m}_{1}x}+{C}_{2}{e}^{{m}_{2}x}$

2.real and equal then the solution is

$y={C}_{1}{e}^{mx}+{C}_{2}x{e}^{mx}$

3.complex then the solution is

$y={e}^{\alpha x}({C}_{1}\mathrm{cos}(\beta x)+{C}_{2}\mathrm{sin}(\beta x))$

Given $y=x{e}^{7x}$ is the solution

the solution is similar for the case of real and equal roots

therefor the other solution is

$y={e}^{7x}$

If roots of auxiliary equation are

1. real and distinct then the solution is

$y={C}_{1}{e}^{{m}_{1}x}+{C}_{2}{e}^{{m}_{2}x}$

2.real and equal then the solution is

$y={C}_{1}{e}^{mx}+{C}_{2}x{e}^{mx}$

3.complex then the solution is

$y={e}^{\alpha x}({C}_{1}\mathrm{cos}(\beta x)+{C}_{2}\mathrm{sin}(\beta x))$

Given $y=x{e}^{7x}$ is the solution

the solution is similar for the case of real and equal roots

therefor the other solution is

$y={e}^{7x}$

asked 2021-10-05

Consider the system of differential equations $\frac{dx}{dt}=-y\text{}\text{}\text{}\frac{dy}{dt}=-x$ .a)Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation.

b)Solve the equation you obtained for y as a function of t; hence find x as a function of t.

b)Solve the equation you obtained for y as a function of t; hence find x as a function of t.

asked 2022-05-28

I would need some help in solving the following differential equation:

${u}^{\prime}(p)+\frac{r+N\lambda p}{N\lambda p(1-p)}u(p)=\frac{r+N\lambda p}{N\lambda (1-p)}g.$

I already know that the solution is of the form

$gp+C(1-p){\left(\frac{1-p}{p}\right)}^{r/\lambda N},$,

where C is a constant.

Now this is a standard linear first-order differential equation. Letting

$f(p)=\frac{r+N\lambda p}{N\lambda p(1-p)},\text{}\text{}q(p)=\frac{r+N\lambda p}{N\lambda (1-p)}g,\text{}\text{}\mu (p)={e}^{\int f(p)dp},$

the solution should have the form

$u(p)=\frac{C}{\mu (p)}+\frac{1}{\mu (p)}\int \mu (p)q(p)dp,$

where C is a constant.

The calculation of $\mu (p)$ was not much of a problem. Indeed, we can rewrite f(p) as

$f(p)=\frac{r}{N\lambda}\frac{1}{p}+(1+\frac{r}{N\lambda})\frac{1}{1-p}$

Using $\int \frac{1}{p}dp=\mathrm{ln}(p)$ and $\int \frac{1}{1-p}dp=-\mathrm{ln}(1-p)$, I found that

$\mu (p)={p}^{\frac{r}{N\lambda}}(1-p{)}^{-1-\frac{r}{N\lambda}}.$

Where I have difficulties is calculating the term ∫μ(p)q(p)dp, where

$\mu (p)q(p)=\frac{rg}{N\lambda}{p}^{\frac{r}{N\lambda}}(1-p{)}^{-2-\frac{r}{N\lambda}}+{p}^{\frac{r}{N\lambda}+1}(1-p{)}^{-2-\frac{r}{N\lambda}}.$

How do I integrate this last term? Or did I miss something obvious, because in the paper I'm looking at they don't provide any details about how they get to the solution of the differential equation

Edit: small typo corrected in the last displayed equation.

Edit: there is a big typo in the differential equation: the rhs should have $r+N\lambda $ rather than $r+N\lambda p$ in the numerator. This probably makes it much simpler to solve. My bad.

${u}^{\prime}(p)+\frac{r+N\lambda p}{N\lambda p(1-p)}u(p)=\frac{r+N\lambda p}{N\lambda (1-p)}g.$

I already know that the solution is of the form

$gp+C(1-p){\left(\frac{1-p}{p}\right)}^{r/\lambda N},$,

where C is a constant.

Now this is a standard linear first-order differential equation. Letting

$f(p)=\frac{r+N\lambda p}{N\lambda p(1-p)},\text{}\text{}q(p)=\frac{r+N\lambda p}{N\lambda (1-p)}g,\text{}\text{}\mu (p)={e}^{\int f(p)dp},$

the solution should have the form

$u(p)=\frac{C}{\mu (p)}+\frac{1}{\mu (p)}\int \mu (p)q(p)dp,$

where C is a constant.

The calculation of $\mu (p)$ was not much of a problem. Indeed, we can rewrite f(p) as

$f(p)=\frac{r}{N\lambda}\frac{1}{p}+(1+\frac{r}{N\lambda})\frac{1}{1-p}$

Using $\int \frac{1}{p}dp=\mathrm{ln}(p)$ and $\int \frac{1}{1-p}dp=-\mathrm{ln}(1-p)$, I found that

$\mu (p)={p}^{\frac{r}{N\lambda}}(1-p{)}^{-1-\frac{r}{N\lambda}}.$

Where I have difficulties is calculating the term ∫μ(p)q(p)dp, where

$\mu (p)q(p)=\frac{rg}{N\lambda}{p}^{\frac{r}{N\lambda}}(1-p{)}^{-2-\frac{r}{N\lambda}}+{p}^{\frac{r}{N\lambda}+1}(1-p{)}^{-2-\frac{r}{N\lambda}}.$

How do I integrate this last term? Or did I miss something obvious, because in the paper I'm looking at they don't provide any details about how they get to the solution of the differential equation

Edit: small typo corrected in the last displayed equation.

Edit: there is a big typo in the differential equation: the rhs should have $r+N\lambda $ rather than $r+N\lambda p$ in the numerator. This probably makes it much simpler to solve. My bad.

asked 2021-12-08

Please suggest a substitution for solving or any method of solving:

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Solve the following IVP using Laplace Transform

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$\left(f\left(x\right)\right)$ -inside function

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${f}^{\prime}\left(x\right)$ -derivative of the inside

Now, explain what happens in regards to calculating the derivative, if f(x) is also a composite function.

Now, explain what happens in regards to calculating the derivative, if f(x) is also a composite function.

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