arenceabigns

2021-03-02

Solve the differential equation $\left(x-3y\right)dx-xdy=0$

Nathanael Webber

$\left(x-3y\right)-x\frac{dy}{dx}=0$
$-\frac{dy}{dx}-\left(\frac{x-3y}{x}\right)=0$
$\frac{dy}{dx}-\frac{x}{x}+\frac{3y}{x}=0$
$\frac{dy}{dx}+\frac{3}{x}y=1$
$\frac{dy}{dx}+Py=Q$
where $P=\frac{3}{x}$, Q=1
$I.F.={e}^{\int Pdx}$
$={e}^{\int \frac{3}{x}dx}$
$={e}^{3\mathrm{ln}abs\left(x\right)}$
$={e}^{\mathrm{ln}\left(x{\right)}^{3}}$
$={x}^{3}$
$y\left(I.F.\right)=\int Q\left(I.F.\right)+C$
$y\left({x}^{3}\right)=\int 1\left({x}^{3}\right)+C$
$=\frac{{x}^{4}}{4}+C$
$y=\frac{x}{4}+\frac{C}{{x}^{3}}$
$=\frac{x}{4}+{C}_{1}$
${C}_{1}=\frac{C}{{x}^{3}}$

Nick Camelot

To solve the differential equation $\left(x-3y\right)dx-xdy=0$, we can use the method of exact differential equations. The equation is exact if we can find a function $u\left(x,y\right)$ such that the total differential $du$ satisfies $\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=0$.
Let's check if the given equation is exact by computing the partial derivatives:
$\frac{\partial }{\partial y}\left(x-3y\right)=-3$
$\frac{\partial }{\partial x}\left(-1\right)=0$
Since the mixed partial derivative $\frac{\partial }{\partial y}\left(-1\right)$ is not equal to $\frac{\partial }{\partial x}\left(-3y\right)$, the equation is not exact. However, we can try to find an integrating factor to make it exact.
We multiply the entire equation by an integrating factor $I\left(x\right)$:
$I\left(x\right)\left(x-3y\right)dx-I\left(x\right)dy=0$
We want the coefficient of $dy$ to be equal to $\frac{\partial u}{\partial y}$. In this case, $\frac{\partial u}{\partial y}=-I\left(x\right)$. We can solve this equation to find $u\left(x,y\right)$.
Integrating $\frac{\partial u}{\partial y}=-I\left(x\right)$ with respect to $y$, we get:
$u\left(x,y\right)=-I\left(x\right)y+g\left(x\right)$
where $g\left(x\right)$ is an arbitrary function of $x$. To determine $I\left(x\right)$, we differentiate $u\left(x,y\right)$ with respect to $x$ and equate it to the coefficient of $dx$:
$\frac{\partial u}{\partial x}=-{I}^{\prime }\left(x\right)y+{g}^{\prime }\left(x\right)=x-3y$
Comparing coefficients, we have:
$-{I}^{\prime }\left(x\right)=1$
${g}^{\prime }\left(x\right)=0$
Solving these equations, we find:
$I\left(x\right)=-x+C$
$g\left(x\right)={C}^{\prime }$
where $C$ and ${C}^{\prime }$ are constants.
Now, we can substitute $I\left(x\right)$ and $g\left(x\right)$ back into $u\left(x,y\right)$:
$u\left(x,y\right)=\left(-x+C\right)y+{C}^{\prime }$
Finally, we equate $u\left(x,y\right)$ to zero since $du=0$:
$\left(-x+C\right)y+{C}^{\prime }=0$

Eliza Beth13

Step 1: Rewrite the equation in the standard form:
$\left(x-3y\right)dx-xdy=0$
Step 2: Divide through by $x$ to separate the variables:
$\frac{\left(x-3y\right)}{x}dx-dy=0$
Step 3: Rearrange the terms:
$\frac{dx}{x}-3\frac{dy}{dx}=0$
Step 4: Introduce a substitution, let $u=\frac{dy}{dx}$:
$\frac{dx}{x}-3udu=0$
Step 5: Integrate both sides of the equation:
$\int \frac{dx}{x}-\int 3udu=\mathrm{ln}|x|-\frac{3{u}^{2}}{2}=C$
Step 6: Substitute back $u=\frac{dy}{dx}$:
$\mathrm{ln}|x|-\frac{3\left(\frac{dy}{dx}{\right)}^{2}}{2}=C$
Step 7: Solve for $y$:
$\frac{dy}{dx}=\sqrt{\frac{2\left(\mathrm{ln}|x|-C\right)}{3}}$
Step 8: Integrate both sides with respect to $x$:
$\int \frac{dy}{\sqrt{\frac{2\left(\mathrm{ln}|x|-C\right)}{3}}}=\int dx$
Step 9: Simplify the integral and solve for $y$:
$y=\int \sqrt{\frac{3}{2}}\frac{dx}{\sqrt{\mathrm{ln}|x|-C}}$
Therefore, the solution to the differential equation $\left(x-3y\right)dx-xdy=0$ is given by $y=\int \sqrt{\frac{3}{2}}\frac{dx}{\sqrt{\mathrm{ln}|x|-C}}$, where $C$ is the constant of integration.

$x-y=3y\mathrm{ln}|x|+C$
Explanation:
The differential equation is:
$\left(x-3y\right)dx-xdy=0$
To begin, we rearrange the terms:
$\left(x-3y\right)dx=xdy$
Now, let's divide both sides by $x$ to separate the variables:
$\frac{\left(x-3y\right)}{x}dx=dy$
Next, we integrate both sides with respect to their respective variables:
$\int \frac{\left(x-3y\right)}{x}dx=\int dy$
To integrate $\int \frac{\left(x-3y\right)}{x}dx$, we can expand the fraction:
$\int \left(1-\frac{3y}{x}\right)dx$
Integrating term by term, we get:
$\int dx-\int \frac{3y}{x}dx$
Integrating the first term with respect to $x$ gives us $x$, and integrating the second term gives us $-3y\mathrm{ln}|x|$:
$x-3y\mathrm{ln}|x|=y+C$
Here, $C$ represents the constant of integration.
Finally, we can rearrange the equation to solve for $y$:
$x-y=3y\mathrm{ln}|x|+C$

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