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Eugene's question via Facebook about a Differential Equation

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Solve $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = 3\,\sqrt{4 - y^2} \end{align*}$ given that $\displaystyle \begin{align*} y\left( 0 \right) = 2 \end{align*}$
This equation is separable...

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}&= 3\,\sqrt{4 - y^2} \\ \frac{1}{\sqrt{4 - y^2}}\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 3 \\ \int{ \frac{1}{\sqrt{4 - y^2}}\,\frac{\mathrm{d}y}{\mathrm{d}x} \,\mathrm{d}x} &= \int{ 3\,\mathrm{d}x} \\ \int{ \frac{1}{\sqrt{4 - y^2}}\,\mathrm{d}y} &= 3\,x + C_1 \end{align*}$

Now let $\displaystyle \begin{align*} y =2\sin{(t)} \implies \mathrm{d}y = 2\cos{(t)}\,\mathrm{d}t \end{align*}$

$\displaystyle \begin{align*} \int{\frac{1}{\sqrt{ 4 - \left[ 2\sin{(t)} \right] ^2} } \,2\cos{(t)} \, \mathrm{d}t } &= 3\,x + C_1 \\ \int{ \frac{ 2\cos{(t)} }{ \sqrt{4 - 4\sin^2{(t)} } }\,\mathrm{d}t} &= 3\,x + C_1 \\ \int{ \frac{2\cos{(t)}}{\sqrt{4\left[ 1 - \sin^2{(t)} \right] } } \,\mathrm{d}t} &= 3\,x + C_1 \\ \int{ \frac{2\cos{(t)}}{\sqrt{4\cos^2{(t)} }}\,\mathrm{d}t} &= 3\,x + C_1 \\ \int{ \frac{2\cos{(t)}}{2\cos{(t)}}\,\mathrm{d}t} &= 3\,x + C_1 \\ \int{1\,\mathrm{d}t} &= 3\,x + C_1 \\ t + C_2 &= 3\,x + C_1 \\ t &= 3\,x + C, \textrm{ where } C = C_1 - C_2 \\ \arcsin{ \left( \frac{y}{2} \right) } &= 3\,x + C \\ \frac{y}{2} &= \sin{ \left( 3\,x + C \right) } \\ y &= 2\sin{ \left( 3\,x + C \right) } \end{align*}$

and since $\displaystyle \begin{align*} y \left( 0 \right) = 2 \end{align*}$

$\displaystyle \begin{align*} 2 &= \sin{ \left[ 3 \left( 0 \right) + C \right] } \\ 2 &= \sin{(C)} \\ C &= \arcsin{ \left( 2 \right) } \end{align*}$

Thus $\displaystyle \begin{align*} y = 2\sin{ \left[ 3\,x + \arcsin{ \left( 2 \right) } \right] } \end{align*}$
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,661
St. Augustine, FL.
...and since $\displaystyle \begin{align*} y \left( 0 \right) = 2 \end{align*}$

$\displaystyle \begin{align*} 2 &= \sin{ \left[ 3 \left( 0 \right) + C \right] } \\ 2 &= \sin{(C)} \\ C &= \arcsin{ \left( 2 \right) } \end{align*}$

Thus $\displaystyle \begin{align*} y = 2\sin{ \left[ 3\,x + \arcsin{ \left( 2 \right) } \right] } \end{align*}$
Just a minor quibble...you want:

\(\displaystyle 2=2\sin(3(0)+C)\implies C=\arcsin(1)\)

Hence:

\(\displaystyle y(x)=2\sin\left(3x+\arcsin(1)\right)\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Just a minor quibble...you want:

\(\displaystyle 2=2\sin(3(0)+C)\implies C=\arcsin(1)\)

Hence:

\(\displaystyle y(x)=2\sin\left(3x+\arcsin(1)\right)\)
This is why I shouldn't tutor at 1am hahaha. And of course, $\displaystyle \begin{align*} \arcsin{(1)} = \frac{\pi}{2} \end{align*}$ :)