Question

$$Y^{\prime\prime}=e^{X}$$

Answer

$$Y=(-e^(X-2))^(1/p)/(r^(2/p)*m^(2/p)*p^(1/p))$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{Y}^{p}{r}^{2}{\imath }^{2}{m}^{2}{e}^{2}p={e}^{X}\]
Use Square Rule: \({i}^{2}=-1\).
\[{Y}^{p}{r}^{2}\times -1\times {m}^{2}{e}^{2}p={e}^{X}\]
Simplify  \({Y}^{p}{r}^{2}\times -1\times {m}^{2}{e}^{2}p\)  to  \({Y}^{p}{r}^{2}\times -{m}^{2}{e}^{2}p\).
\[{Y}^{p}{r}^{2}\times -{m}^{2}{e}^{2}p={e}^{X}\]
Regroup terms.
\[-{e}^{2}{Y}^{p}{r}^{2}{m}^{2}p={e}^{X}\]
Divide both sides by \(-{e}^{2}\).
\[{Y}^{p}{r}^{2}{m}^{2}p=-\frac{{e}^{X}}{{e}^{2}}\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[{Y}^{p}{r}^{2}{m}^{2}p=-{e}^{X-2}\]
Divide both sides by \({r}^{2}\).
\[{Y}^{p}{m}^{2}p=-\frac{{e}^{X-2}}{{r}^{2}}\]
Divide both sides by \({m}^{2}\).
\[{Y}^{p}p=-\frac{\frac{{e}^{X-2}}{{r}^{2}}}{{m}^{2}}\]
Simplify  \(\frac{\frac{{e}^{X-2}}{{r}^{2}}}{{m}^{2}}\)  to  \(\frac{{e}^{X-2}}{{r}^{2}{m}^{2}}\).
\[{Y}^{p}p=-\frac{{e}^{X-2}}{{r}^{2}{m}^{2}}\]
Divide both sides by \(p\).
\[{Y}^{p}=-\frac{\frac{{e}^{X-2}}{{r}^{2}{m}^{2}}}{p}\]
Simplify  \(\frac{\frac{{e}^{X-2}}{{r}^{2}{m}^{2}}}{p}\)  to  \(\frac{{e}^{X-2}}{{r}^{2}{m}^{2}p}\).
\[{Y}^{p}=-\frac{{e}^{X-2}}{{r}^{2}{m}^{2}p}\]
Take the \((p)\)th root of both sides.
\[Y=\sqrt[p]{-\frac{{e}^{X-2}}{{r}^{2}{m}^{2}p}}\]
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[Y=\frac{\sqrt[p]{-{e}^{X-2}}}{\sqrt[p]{{r}^{2}{m}^{2}p}}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[Y=\frac{\sqrt[p]{-{e}^{X-2}}}{\sqrt[p]{{r}^{2}}\sqrt[p]{{m}^{2}}\sqrt[p]{p}}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[Y=\frac{\sqrt[p]{-{e}^{X-2}}}{{r}^{\frac{2}{p}}\sqrt[p]{{m}^{2}}\sqrt[p]{p}}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[Y=\frac{\sqrt[p]{-{e}^{X-2}}}{{r}^{\frac{2}{p}}{m}^{\frac{2}{p}}\sqrt[p]{p}}\]