$x{y}^{\prime}={y}^{2}-(2x+1)y+{x}^{2}+2x$

where $y=ax+b$ is a particular solution.

Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

Ellen Chang
2022-07-13
Answered

What is the standard method for finding solutions of differential equations such as this one? (if there is any)

$x{y}^{\prime}={y}^{2}-(2x+1)y+{x}^{2}+2x$

where $y=ax+b$ is a particular solution.

Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

$x{y}^{\prime}={y}^{2}-(2x+1)y+{x}^{2}+2x$

where $y=ax+b$ is a particular solution.

Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

You can still ask an expert for help

bap1287dg

Answered 2022-07-14
Author has **13** answers

I think that your hint is "there exists $a,b$ such thaht $y(x)=ax+b$ is solution". We find easily that $y(x)=x$ and $y(x)=x+1$ are solutions.

Hint: Note that your equation is

$x{y}^{\mathrm{\prime}}(x)-x=(y(x)-x)(y(x)-x-1)$

Now put $y(x)=x+z(x)$.

Hint: Note that your equation is

$x{y}^{\mathrm{\prime}}(x)-x=(y(x)-x)(y(x)-x-1)$

Now put $y(x)=x+z(x)$.

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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?

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Differential equations in the following form are called Bernoulli Equations.

Find the solution for the following initial value problems and find the interval of validity for the solution.

${y}^{\prime}+\frac{4}{x}y={x}^{3}{y}^{2}\text{}y\left(2\right)=-1,x0$

Find the solution for the following initial value problems and find the interval of validity for the solution.

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$\left({d}^{2}\frac{y}{d{t}^{2}}\right)\text{}+\text{}7\left(\frac{dy}{dt}\right)\text{}+\text{}10y=4t{e}^{-3}t$ with

$y(0)=0,\text{}{y}^{\prime}(0)=\text{}-1$

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$c\left(t\right)=(t\mathrm{sin}t,4t)$

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