Jason Farmer
2021-02-24
Answered

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

You can still ask an expert for help

krolaniaN

Answered 2021-02-25
Author has **86** answers

Divide both sides by x and further simplify it

Substitute

Differentiate with respect to x

Hence,

Substract v from both sides and further simplify it

Integrate both sides

Taking log both sides

asked 2021-11-19

Differentiate.

$y=x{\mathrm{cos}}^{-1}x$

asked 2021-09-07

Use Laplace transform to solve the initial-value problem

asked 2022-06-24

Given the first-order differential equation:

$\frac{dy}{dx}=-6xy$

The textbook says you cannot differentiate both sides as y is on the right side and you have to use separation of variables. However, I did integrate both sides and arrived at this:

$\int \frac{dy}{dx}dx=\int -6xydx$

$y=-6y\int xdx$

$y=-6y(\frac{{x}^{2}}{2}+C)$

$y=-3y{x}^{2}+C$

$y+3y{x}^{2}=C$

$y(1+3{x}^{2})=C$

$y(x)=\frac{C}{1+{x}^{2}}$

And the second last step is valid since $1+3{x}^{2}$ can never be zero. However, this is not the correct answer, which is:

$y(x)=C{e}^{-3{x}^{2}}$. Why is this?

$\frac{dy}{dx}=-6xy$

The textbook says you cannot differentiate both sides as y is on the right side and you have to use separation of variables. However, I did integrate both sides and arrived at this:

$\int \frac{dy}{dx}dx=\int -6xydx$

$y=-6y\int xdx$

$y=-6y(\frac{{x}^{2}}{2}+C)$

$y=-3y{x}^{2}+C$

$y+3y{x}^{2}=C$

$y(1+3{x}^{2})=C$

$y(x)=\frac{C}{1+{x}^{2}}$

And the second last step is valid since $1+3{x}^{2}$ can never be zero. However, this is not the correct answer, which is:

$y(x)=C{e}^{-3{x}^{2}}$. Why is this?

asked 2022-01-20

Solve and find the soltion to the first order differential equation:

$x\frac{dy}{dx}+y={e}^{x},\text{}\text{}x0$

asked 2021-09-12

Find the inverse Laplace Transform for the following expressions

$\frac{5s}{{(s-2)}^{2}}$

asked 2021-09-06

Find the inverse Laplace transformation for $\frac{2s-11}{({s}^{2}-4s+8)}$

asked 2022-01-19

Solve: $\frac{dp}{dt}={t}^{2}p-p+{t}^{2}-1$

I got$\int \frac{1}{p+1}dp=\int ({t}^{2}-1)dt$

After take the integral I got

$\mathrm{ln}|p+1|+c=\frac{1}{3}{t}^{2}-t+c$

after this step I stacked. How can I simplify like$p=$ something?

I got

After take the integral I got

after this step I stacked. How can I simplify like