What is a solution to the differential equation $\frac{dy}{dx}=x+y$?

Joyce Sharp
2022-09-18
Answered

What is a solution to the differential equation $\frac{dy}{dx}=x+y$?

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asked 2022-01-20

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Solve the following first order differential equations. Use any method:

$2{x}^{2}dx-3x{y}^{2}dy=0;\text{}y\left(1\right)=0$

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Solve differential equation$x{y}^{\prime}+[(2x+1)/(x+1)]y=x-1$

asked 2022-06-12

Let the first order differential equation be

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$

asked 2020-11-07

Solve differential equation
$xydy-{y}^{2}dx=(x+y{)}^{2}{e}^{(}-y/x)$

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 2022-05-22

$t{y}^{\prime}+ty=1-y$

I am having trouble going about this question. I have tried different processes to answer the differential equation, including separable, exact, homogeneous and linear equations. I have ran into a wall through all these processes. Maybe I'm not following the correct procedure... Help!

I know the first step would be to divide both sides by t, which would give:

${y}^{\prime}+y=(1-y)/t$

I am having trouble going about this question. I have tried different processes to answer the differential equation, including separable, exact, homogeneous and linear equations. I have ran into a wall through all these processes. Maybe I'm not following the correct procedure... Help!

I know the first step would be to divide both sides by t, which would give:

${y}^{\prime}+y=(1-y)/t$

asked 2021-02-24

Solve the differential equation
$xdy/dx=y+x{e}^{(}y/x)$ , y=vx