gorovogpg

2021-12-30

Exact Differential Equations Solve $\left(2x+\mathrm{sin}y\right)dx+\left(x\mathrm{cos}y+4{y}^{3}\right)dy=0$

veiga34

$\left(2x+\mathrm{sin}y\right)dx+\left(x\mathrm{ln}y+4{y}^{3}\right)dy=0$

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
$\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left(2x+\mathrm{sin}y\right)=\mathrm{ln}y$
$\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left(x\mathrm{ln}y+4{y}^{3}\right)=\mathrm{ln}y$
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
Hence given DE is exact
$\int \left(2x+\mathrm{sin}y\right)dx+\int 4{y}^{3}dy$
$\int 2xdx+\mathrm{sin}y\int dx+4\int {y}^{3}dy=c$
$\alpha ×\frac{{x}^{2}}{2}+\mathrm{sin}y\left(x\right)+4×\frac{{y}^{4}}{4}$
${x}^{2}+x\mathrm{sin}y+{y}^{4}=c$

Paul Mitchell

$⇒2xdx+\left(\mathrm{sin}ydx+x\mathrm{cos}ydy\right)+4{y}^{3}dy=0$
$⇒\alpha \left({x}^{2}\right)+\alpha \left(x\mathrm{sin}y\right)+\alpha \left({y}^{4}\right)=0$
$⇒\alpha \left({x}^{2}+x\mathrm{sin}y+{y}^{4}\right)=0$
Integrating both side we get,
${x}^{2}+x\mathrm{sin}y+{y}^{4}=c$
Hence the required solution is
${x}^{2}+x\mathrm{sin}y+{y}^{4}=c$

karton

$\begin{array}{}⇒2xdx+\left(\mathrm{sin}ydx+x\mathrm{cos}ydy\right)+4{y}^{3}dy=0\\ ⇒\alpha \left({x}^{2}\right)+\alpha \left(x\mathrm{sin}y\right)+\alpha \left({y}^{4}\right)=0\\ ⇒\alpha \left({x}^{2}+x\mathrm{sin}y+{y}^{4}\right)=0\\ {x}^{2}+x\mathrm{sin}y+{y}^{4}=c\\ {x}^{2}+x\mathrm{sin}y+{y}^{4}=c\end{array}$