Donald Johnson

2021-12-21

$\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$

Stella Calderon

The given differential equation is
$\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$
Comparing it with $Mdx+Mdy=0$

Here, $\frac{dM}{dy}=\frac{dN}{dx}$
Then, it is exact differential equation, so solution is
$\int Mdx+\int N$ (term not contining x) $dy=c$
$\int \left(2x-3{x}^{2}\right)dx+\int ydy=c$
$y\cdot \frac{2{x}^{2}}{2}-\frac{3{x}^{3}}{3}+\frac{{y}^{2}}{2}=c$
${x}^{2}y-{x}^{3}+\frac{{y}^{2}}{2}=c$
This is solution of given DE.

Charles Benedict

Assuming $\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$
Then $2xy+\left({x}^{2}+y\right)\frac{dy}{dx}=3{x}^{2}$ (1)
Notice that $d\frac{{x}^{2}y+\frac{{y}^{2}}{2}}{dx}$ is:
${x}^{2}\frac{dy}{dx}+2xy+2\frac{y}{2}\frac{dy}{dx}$
$=2xy+\left({x}^{2}+y\right)\frac{dy}{dx}$ which is left hand side of (1)
So integrating (1) gives
${x}^{2}y+\frac{{y}^{2}}{2}={x}^{3}+k$
So ${y}^{2}+2{x}^{2}y-2{x}^{3}+c=0$ is the solution

RizerMix

Let $M=\left(2xy-3{x}^{2}\right)$ so that $DM/Dy=2x$
Let $N=:\left({x}^{2}+y\right)$ so that $DN/Dx=2x$
since $DM/Dy=DN/Dx$ here D is to be read as delta.
The given Equation
$\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$ is exact
Integrating $y{x}^{2}x-6x+1/2{y}^{2}=0$
or, $2y{x}^{2}x-12x+{y}^{2}=0$

Nick Camelot

Result:
${x}^{2}y-{x}^{3}+C=0.$
Solution:
First, we check if the equation is exact by verifying if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, where $M=2xy-3{x}^{2}$ and $N={x}^{2}+y$.
Calculating the partial derivatives:
$\frac{\partial M}{\partial y}=2x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial N}{\partial x}=2x,$
Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, the equation is exact.
To find the solution, we integrate $M$ with respect to $x$ while treating $y$ as a constant, and then find the potential function $\varphi \left(x,y\right)$:
$\varphi \left(x,y\right)=\int \left(2xy-3{x}^{2}\right)dx={x}^{2}y-{x}^{3}+f\left(y\right),$
where $f\left(y\right)$ is an arbitrary function of $y$.
Next, we differentiate $\varphi \left(x,y\right)$ with respect to $y$ and equate it to $N$:
$\frac{\partial \varphi }{\partial y}={x}^{2}+\frac{df}{dy}={x}^{2}+y=N.$
Comparing the equation above with $N={x}^{2}+y$, we can see that $\frac{df}{dy}=0$.
Integrating $\frac{df}{dy}=0$ with respect to $y$, we obtain $f\left(y\right)=C$, where $C$ is a constant.
Therefore, the potential function $\varphi \left(x,y\right)$ is:
$\varphi \left(x,y\right)={x}^{2}y-{x}^{3}+C.$
To find the final solution, we set $\varphi \left(x,y\right)$ equal to a constant, which gives us:
${x}^{2}y-{x}^{3}+C=0.$

Mr Solver

Step 1: Identify the equation
The given differential equation is:
$\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$
Step 2: Separate the variables
To solve the equation, we need to separate the variables $x$ and $y$.
$\left(2xy-3{x}^{2}\right)dx=-\left({x}^{2}+y\right)dy$
Step 3: Integrate both sides
Now, we integrate both sides of the equation.
$\int \left(2xy-3{x}^{2}\right)\phantom{\rule{0.167em}{0ex}}dx=-\int \left({x}^{2}+y\right)\phantom{\rule{0.167em}{0ex}}dy$
Step 4: Evaluate the integrals
Integrating the left side with respect to $x$ gives:
$\int 2xy-3{x}^{2}\phantom{\rule{0.167em}{0ex}}dx=2\int xy\phantom{\rule{0.167em}{0ex}}dx-3\int {x}^{2}\phantom{\rule{0.167em}{0ex}}dx$
Using the power rule for integration, we have:
$=2\int xy\phantom{\rule{0.167em}{0ex}}dx-3\int {x}^{2}\phantom{\rule{0.167em}{0ex}}dx$
$=2\left(\frac{1}{2}x{y}^{2}\right)-3\left(\frac{1}{3}{x}^{3}\right)+{C}_{1}$
where ${C}_{1}$ is the constant of integration.
Similarly, integrating the right side with respect to $y$ gives:
$-\int \left({x}^{2}+y\right)\phantom{\rule{0.167em}{0ex}}dy=-\int {x}^{2}\phantom{\rule{0.167em}{0ex}}dy-\int y\phantom{\rule{0.167em}{0ex}}dy$
$=-\frac{1}{3}{x}^{3}-\frac{1}{2}{y}^{2}+{C}_{2}$
where ${C}_{2}$ is another constant of integration.
Step 5: Equate the integrals
Setting the integrals equal to each other, we have:
$2\left(\frac{1}{2}x{y}^{2}\right)-3\left(\frac{1}{3}{x}^{3}\right)+{C}_{1}=-\frac{1}{3}{x}^{3}-\frac{1}{2}{y}^{2}+{C}_{2}$
Simplifying the equation, we get:
$x{y}^{2}-{x}^{3}+{C}_{1}=-\frac{1}{3}{x}^{3}-\frac{1}{2}{y}^{2}+{C}_{2}$
Step 6: Combine the constants
Combining the constants ${C}_{1}$ and ${C}_{2}$ into a single constant $C$, we have:
$x{y}^{2}-{x}^{3}+C=-\frac{1}{3}{x}^{3}-\frac{1}{2}{y}^{2}$
Rearranging the equation, we obtain the solution:
$x{y}^{2}-{x}^{3}+\frac{1}{3}{x}^{3}+\frac{1}{2}{y}^{2}+C=0$
Simplifying further, we get:
$x{y}^{2}+\frac{2}{3}{x}^{3}+\frac{1}{2}{y}^{2}+C=0$
Therefore, the solution to the given differential equation is:
$x{y}^{2}+\frac{2}{3}{x}^{3}+\frac{1}{2}{y}^{2}+C=0$
where $C$ is the constant of integration.

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