 namenerk

2021-12-30

$\left(2{y}^{2}-2xy+3x\right)dx+\left(y+4xy-{x}^{2}\right)dy=0$ twineg4

$\left({e}^{x}\mathrm{sin}y+2y\right)dx+\left(2x-{e}^{x}\mathrm{sin}y\right)dy=0$
$\frac{\partial p}{\partial y}={e}^{x}\mathrm{cos}y+2$
$\frac{\partial Q}{\partial x}=2-{e}^{x}\mathrm{sin}y$
$\therefore \frac{\partial p}{\partial y}\ne \frac{\partial Q}{\partial x}$
The differential equation is not exact. Linda Birchfield

This is an exact differential equation, because,

$D\frac{f}{D}y=4xy$
$f\left(x,y\right)=2x{y}^{2}+G\left(x\right)$
$Dxf\left(x,y\right)={\left(xy\right)}^{2}+{G}^{\prime }\left(x\right)$
Now,
${\left(Xy\right)}^{2}+{G}^{\prime }\left(x\right)=2{y}^{2}+3x$
Solving for G(x), we have
which is:
$3\frac{{x}^{2}}{2}+2x{y}^{2}-\frac{{x}^{4}{y}^{2}}{2}+k$. …A
General solution becomes,
$A+2x{y}^{2}$ karton

Solve for $\left(dy\left(x\right)\right)/\left(dx\right)$:
$\left(dy\left(x\right)\right)/\left(dx\right)=\left(-x+4y\left(x\right)+9\right)/\left(4x+y\left(x\right)-2\right)$
Let . This gives :
$\left(dv\left(t\right)\right)/\left(dt\right)=\left(-t+4\left(v\left(t\right)-2\right)+8\right)/\left(4\left(t+1\right)+v\left(t\right)-4\right)$
Collect in terms of t and v(t):
$\left(dv\left(t\right)\right)/\left(dt\right)=\left(-t+4v\left(t\right)\right)/\left(4t+v\left(t\right)\right)$
Let v(t) = t u(t), which gives $\left(dv\left(t\right)\right)/\left(dt\right)=t\left(du\left(t\right)\right)/\left(dt\right)+u\left(t\right):$
$t\left(du\left(t\right)\right)/\left(dt\right)+u\left(t\right)=\left(-t+4tu\left(t\right)\right)/\left(4t+tu\left(t\right)\right)$
Simplify:
$t\left(du\left(t\right)\right)/\left(dt\right)+u\left(t\right)=\left(4u\left(t\right)-1\right)/\left(u\left(t\right)+4\right)$
Solve for $\left(du\left(t\right)\right)/\left(dt\right)$:
$\left(du\left(t\right)\right)/\left(dt\right)=\left(-u\left(t{\right)}^{2}-1\right)/\left(t\left(u\left(t\right)+4\right)\right)$
Divide both sides by $\left(-u\left(t{\right)}^{2}-1\right)/\left(u\left(t\right)+4\right)$:
$\left(\left(du\left(t\right)\right)/\left(dt\right)\left(u\left(t\right)+4\right)\right)/\left(-u\left(t{\right)}^{2}-1\right)=1/t$
Integrate both sides with respect to t:
integral $\left(\left(du\left(t\right)\right)/\left(dt\right)\left(u\left(t\right)+4\right)\right)/\left(-u\left(t{\right)}^{2}-1\right)dt=\text{integral}1/tdt$
Evaluate the integrals:
$-4ta{n}^{-1}\left(u\left(t\right)\right)-1/2\mathrm{log}\left(u\left(t{\right)}^{2}+1\right)=\mathrm{log}\left(t\right)+{c}_{1}$, where ${c}_{1}$ is an arbitrary constant.
Substitute back for v(t) = t u(t):
$-4ta{n}^{-1}\left(v\left(t\right)/t\right)-1/2\mathrm{log}\left(v\left(t{\right)}^{2}/{t}^{2}+1\right)=\mathrm{log}\left(t\right)+{c}_{1}$
Substitute back for :
Answer: $-4ta{n}^{-1}\left(\left(y\left(x\right)+2\right)/\left(x-1\right)\right)-1/2\mathrm{log}\left(\left(y\left(x\right)+2{\right)}^{2}/\left(x-1{\right)}^{2}+1\right)=\mathrm{log}\left(x-1\right)+{c}_{1}$

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