# Thread: Hard differential equation problem

1. ## Hard differential equation problem

Solve y'= sin(4x+5y)
Appreciate any help.

2. ## Re: Hard differential equation problem

\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \sin{ \left( 4\,x + 5\,y \right) } \\ 5\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 5\sin{ \left( 4\,x + 5\,y \right) } \\ 4 + 5\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 4 + 5\sin{ \left( 4\,x + 5\,y \right) } \end{align*}

Let \displaystyle \begin{align*} u = 4\,x + 5\,y \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 4 + 5\,\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*} and the DE becomes

\displaystyle \begin{align*} 4 + 5\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 4 + 5\sin{ \left( 4\,x + 5\,y \right) } \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= 4 + 5\sin{ \left( u \right) } \\ \left[ \frac{1}{4 + 5\sin{ \left( u \right) } } \right] \frac{\mathrm{d}u}{\mathrm{d}x} &= 1 \\ \int{ \left[ \frac{1}{4 + 5\sin{ \left( u \right) }} \right] \frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x } &= \int{ 1\,\mathrm{d}x} \\ \int{ \left[ \frac{1}{4 + 5\sin{ \left( u \right) }} \right] \mathrm{d}u } &= x + C_1 \end{align*}

Now let

\displaystyle \begin{align*} t &= \tan{ \left( \frac{u}{2} \right) } \\ \mathrm{d}t &= \frac{1}{2} \, \sec^2{ \left( \frac{u}{2} \right) } \,\mathrm{d}u \\ 2\,\mathrm{d}t &= \left[ 1 + \tan^2{ \left( \frac{u}{2} \right) } \right] \mathrm{d}u \\ 2\,\mathrm{d}t &= \left( 1 + t^2 \right) \mathrm{d}u \\ \left( \frac{2}{1 + t^2} \right) \mathrm{d}t &= \mathrm{d}u \end{align*}

and then

\displaystyle \begin{align*} \sin{ \left( u \right) } &= 2\sin{ \left( \frac{u}{2} \right) } \cos{ \left( \frac{u}{2} \right) } \\ &= \frac{ 2 \tan{ \left( \frac{u}{2} \right) }}{\sec^2{ \left( \frac{u}{2} \right) }} \\ &= \frac{2\tan{ \left( \frac{u}{2} \right) }}{1 + \tan^2{ \left( \frac{u}{2} \right) }} \\ &= \frac{2\,t}{1 + t^2} \end{align*}

so the DE becomes

\displaystyle \begin{align*} \int{ \left[ \frac{1}{4 + 5\sin{ \left( u \right) } } \right] \mathrm{d}u } &= x + C_1 \\ \int{ \frac{1}{4 + 5 \left( \frac{2\,t}{1 + t^2} \right) } \left( \frac{2}{1 + t^2} \right) \mathrm{d}t } &= x + C_1 \\ \int{ \left[ \frac{1}{2\left( 1 + t^2 \right) + 5\,t } \right] \mathrm{d}t } &= x + C_1 \\ \int{ \left( \frac{1}{2\,t^2 + 5\,t + 2} \right) \mathrm{d}t } &= x + C_1 \\ \int{ \frac{1}{\left( 2\,t + 1 \right) \left( t + 2 \right) } \,\mathrm{d}t } &= x + C_1 \end{align*}

Now apply Partial Fractions

\displaystyle \begin{align*} \frac{A}{2\,t + 1} + \frac{B}{t + 2} &\equiv \frac{1}{\left( 2\,t + 1 \right) \left( t + 2 \right) } \\ A \left( t + 2 \right) + B \left( 2\,t + 1 \right) &\equiv 1 \end{align*}

Let \displaystyle \begin{align*} t = -2 \end{align*} to find \displaystyle \begin{align*} -3\,B = 1 \implies B = -\frac{1}{3} \end{align*}

Let \displaystyle \begin{align*} t = -\frac{1}{2} \end{align*} to find \displaystyle \begin{align*} \frac{3}{2}\,A = 1 \implies A = \frac{2}{3} \end{align*}

so

\displaystyle \begin{align*} \int{ \left( \frac{\frac{2}{3}}{2\,t + 1} - \frac{\frac{1}{3}}{t + 2} \right) \mathrm{d}t } &= x + C_1 \\ \frac{1}{3} \, \ln{ \left| 2\,t + 1 \right| } - \frac{1}{3} \, \ln{ \left| t + 2 \right| } + C_2 &= x + C_1 \\ \frac{1}{3} \, \ln{ \left| \frac{2\,t + 1}{t + 2} \right| } &= x + C_1 - C_2 \\ \ln{ \left| \frac{2\,t + 1}{t + 2} \right| } &= 3\,x + 3\,C_1 - 3\,C_2 \\ \left| \frac{2\,t + 1}{t + 2} \right| &= \mathrm{e}^{3\,x + 3\,C_1 - 3\,C_2} \\ \left| \frac{2\,t + 1}{t + 2} \right| &= \mathrm{e}^{3\,C_1 - 3\,C_2} \,\mathrm{e}^{3\,x} \\ \frac{2\,t + 1}{t + 2} &= C\,\mathrm{e}^{3\,x} \textrm{ where } C = \pm \mathrm{e}^{3\,C_1 - 3\,C_2} \\ 2 - \frac{3}{t + 2} &= C\,\mathrm{e}^{3\,x} \\ 2 - C\,\mathrm{e}^{3\,x} &= \frac{3}{t + 2} \\ 2 - C\,\mathrm{e}^{3\,x} &= \frac{3}{\tan{ \left( \frac{u}{2} \right) } + 2 } \\ 2 - C\,\mathrm{e}^{3\,x} &= \frac{3}{ \tan{ \left( \frac{4\,x + 5\,y}{2} \right) } + 2 } \\ \left[ \tan{ \left( \frac{4\,x + 5\,y}{2} \right) } +2 \right] \left( 2 - C\,\mathrm{e}^{3\,x} \right) &= 3 \\ \tan{ \left( \frac{4\,x + 5\,y}{2} \right) } + 2 &= \frac{3}{2 - C\,\mathrm{e}^{3\,x} } \\ \tan{ \left( \frac{4\,x + 5\,y}{2} \right) } &= \frac{3}{2 - C\,\mathrm{e}^{3\,x} } - 2 \\ \tan{ \left( \frac{4\,x + 5\,y}{2} \right) } &= \frac{ 2\,C\,\mathrm{e}^{3\,x} - 1 }{2 - C\,\mathrm{e}^{3\,x}} \\ \frac{4\,x + 5\,y}{2} &= \arctan{ \left( \frac{2\,C \,\mathrm{e}^{3\,x} - 1}{2 - C\,\mathrm{e}^{3\,x}} \right) } \\ 4\,x + 5\,y &= 2\arctan{ \left( \frac{2\,C\,\mathrm{e}^{3\,x} - 1}{2 - C\,\mathrm{e}^{3\,x}} \right) } \\ 5\,y &= 2\arctan{ \left( \frac{2\,C\,\mathrm{e}^{3\,x} - 1}{2 - C\,\mathrm{e}^{3\,x}} \right) } - 4\,x \\ y &= \frac{1}{5} \left[ 2\arctan{ \left( \frac{2\,C\,\mathrm{e}^{3\,x} - 1}{2 - C\,\mathrm{e}^{3\,x}} \right) } - 4\,x \right] \end{align*}

3. ## Re: Hard differential equation problem

Thank you, thank you, and thank you.

4. ## Re: Hard differential equation problem

Correct me if I am wrong, please.
At $\displaystyle \int \dfrac{\frac{2}{3}}{2t+1}-\dfrac{\frac{1}{3}}{t+2}dt=x+C_1$

The next line should be

$\displaystyle ln|2t+1|^{2/3}-ln|t+2|^{1/3}=x+C_2$

They cannot be subtracted because the value inside of the ln are not the same. Then, it should be

$\displaystyle ln\sqrt[3]{\dfrac{(2t+1)^2}{t+2}}=x+C_2$

or $\displaystyle \left|\dfrac{(2t+1)^2}{t+2}\right|=e^{3x+C_2}$

And then, the answer is different from the previous one. Am I right?

5. ## Re: Hard differential equation problem

Originally Posted by Loser66
Correct me if I am wrong, please.
At $\displaystyle \int \dfrac{\frac{2}{3}}{2t+1}-\dfrac{\frac{1}{3}}{t+2}dt=x+C_1$

The next line should be

$\displaystyle ln|2t+1|^{2/3}-ln|t+2|^{1/3}=x+C_2$

They cannot be subtracted because the value inside of the ln are not the same. Then, it should be

$\displaystyle ln\sqrt[3]{\dfrac{(2t+1)^2}{t+2}}=x+C_2$
No.

$\displaystyle \int \dfrac{\tfrac{2}{3}}{2t+1}dt = \dfrac{2}{3}\dfrac{1}{2}\ln |2t+1| = \dfrac{1}{3} \ln |2t+1|$ as Prove It showed.