Lahenda e^y(1+x^2)dy=2x(1

$$e ^ { y } ( 1 + x ^ { 2 } ) d y = 2 x ( 1$$

Lahendage ja leidke d (complex solution)

$$\left\{\begin{matrix}d=\frac{2x}{y\left(x^{2}+1\right)e^{y}}\text{, }&x\neq -i\text{ and }x\neq i\text{ and }y\neq 0\\d\in \mathrm{C}\text{, }&x=0\text{ and }y=0\end{matrix}\right,$$

Juhised lineaarvõrrandi lahendamiseks

Kasutage distributiivsusomadust, et korrutada e^{y} ja 1+x^{2}.

$$\left(e^{y}+e^{y}x^{2}\right)dy=2x\times 1$$

Kasutage distributiivsusomadust, et korrutada e^{y}+e^{y}x^{2} ja d.

$$\left(e^{y}d+e^{y}x^{2}d\right)y=2x\times 1$$

Kasutage distributiivsusomadust, et korrutada e^{y}d+e^{y}x^{2}d ja y.

$$e^{y}dy+e^{y}x^{2}dy=2x\times 1$$

Korrutage 2 ja 1, et leida 2.

$$e^{y}dy+e^{y}x^{2}dy=2x$$

Kombineerige kõik liikmed, mis sisaldavad d.

$$\left(e^{y}y+e^{y}x^{2}y\right)d=2x$$

Võrrand on standardkujul.

$$\left(yx^{2}e^{y}+ye^{y}\right)d=2x$$

Jagage mõlemad pooled e^{y}x^{2}y+e^{y}y-ga.

$$\frac{\left(yx^{2}e^{y}+ye^{y}\right)d}{yx^{2}e^{y}+ye^{y}}=\frac{2x}{yx^{2}e^{y}+ye^{y}}$$

e^{y}x^{2}y+e^{y}y-ga jagamine võtab e^{y}x^{2}y+e^{y}y-ga korrutamise tagasi.

$$d=\frac{2x}{yx^{2}e^{y}+ye^{y}}$$

Jagage 2x väärtusega e^{y}x^{2}y+e^{y}y.

$$d=\frac{2x}{y\left(x^{2}+1\right)e^{y}}$$

Lahendage ja leidke d

$$\left\{\begin{matrix}d=\frac{2x}{y\left(x^{2}+1\right)e^{y}}\text{, }&y\neq 0\\d\in \mathrm{R}\text{, }&x=0\text{ and }y=0\end{matrix}\right,$$

Juhised lineaarvõrrandi lahendamiseks

Kasutage distributiivsusomadust, et korrutada e^{y} ja 1+x^{2}.

$$\left(e^{y}+e^{y}x^{2}\right)dy=2x\times 1$$

Kasutage distributiivsusomadust, et korrutada e^{y}+e^{y}x^{2} ja d.

$$\left(e^{y}d+e^{y}x^{2}d\right)y=2x\times 1$$

Kasutage distributiivsusomadust, et korrutada e^{y}d+e^{y}x^{2}d ja y.

$$e^{y}dy+e^{y}x^{2}dy=2x\times 1$$

Korrutage 2 ja 1, et leida 2.

$$e^{y}dy+e^{y}x^{2}dy=2x$$

Kombineerige kõik liikmed, mis sisaldavad d.

$$\left(e^{y}y+e^{y}x^{2}y\right)d=2x$$

Võrrand on standardkujul.

$$\left(yx^{2}e^{y}+ye^{y}\right)d=2x$$

Jagage mõlemad pooled e^{y}x^{2}y+e^{y}y-ga.

$$\frac{\left(yx^{2}e^{y}+ye^{y}\right)d}{yx^{2}e^{y}+ye^{y}}=\frac{2x}{yx^{2}e^{y}+ye^{y}}$$

e^{y}x^{2}y+e^{y}y-ga jagamine võtab e^{y}x^{2}y+e^{y}y-ga korrutamise tagasi.

$$d=\frac{2x}{yx^{2}e^{y}+ye^{y}}$$

Jagage 2x väärtusega e^{y}x^{2}y+e^{y}y.

$$d=\frac{2x}{y\left(x^{2}+1\right)e^{y}}$$

Lahendage ja leidke x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{1-\left(dy\right)^{2}e^{2y}}+1}{dye^{y}}\text{; }x=\frac{-\sqrt{1-\left(dy\right)^{2}e^{2y}}+1}{dye^{y}}\text{, }&d\neq 0\text{ and }y\neq 0\\x=\frac{dye^{y}}{2}\text{, }&d=0\text{ or }y=0\end{matrix}\right,$$

Lahendage ja leidke x

$$\left\{\begin{matrix}x=\frac{\sqrt{1-\left(dy\right)^{2}e^{2y}}+1}{dye^{y}}\text{; }x=\frac{-\sqrt{1-\left(dy\right)^{2}e^{2y}}+1}{dye^{y}}\text{, }&d\neq 0\text{ and }y\neq 0\text{ and }|d|\leq \frac{1}{|y|e^{y}}\\x=\frac{dye^{y}}{2}\text{, }&d=0\text{ or }y=0\end{matrix}\right,$$