Question

solve the following differential equation first order

Answer 1

\[dy \div (dx)+2(\tan2x)y=3e ^{3x}timescos2x \]

Answer 2

\[\frac{ dy }{ dx }+2(\tan 2x)y=3 e ^{3x}\cos 2x\]

Answer 3

right , i just want to know if we are going to use method of linear equation , and to check my answer

Answer 4

yes

Answer 5

multiply by an integration factor \(\mu\) to turn the left-hand side into \(\mu\dfrac{dy}{dx}+\dfrac{d\mu}{dx}y=\dfrac{d}{dx}(\mu y)\)

Answer 6

okay and then get the formula will be ... y g.s = 1/mu (integration of two functions multiplay each other ??

Answer 7

in other words we want $$\frac{d\mu}{dx}=2\tan(2x)\mu$$which solves easily using separation of variables:$$\frac1\mu\frac{d\mu}{dx}=2\tan(2x)\\\int\frac1\mu \frac{d\mu}{dx}dx=\int2\tan(2x)\ dx\\\int\frac1\mu d\mu=\int2\tan(2x)\ dx\\\log \mu=\log\sec(2x)\\\mu=\sec(2x)$$

Answer 8

using the fact that$$\int\tan x\ dx=\log(\sec(x))+C$$

Answer 9

multiply both sides by \(\mu=\sec(2x)\) and we get:$$\frac{d}{dx}(\mu y)=3e^{3x}\cos(2x)\sec(2x)$$as expected; notice \(\cos(2x)\sec(2x)=1\) hence$$\frac{d}{dx}(\mu y)=3e^{3x}\\\int\frac{d}{dx}(\mu y)\ dx=\int3e^{3x}\ dx\\\mu y=e^{3x}+C\\\sec(2x)y=e^{3x}+C\\y=e^{3x}\cos(2x)+C\cos(2x)$$

Answer 10

is it okay to say if \[\int\limits tanx=lnsecx .. & \int\limits \tan2x=lnsec2x/2 ??\]

Answer 11

$$\int\tan(x)\ dx=\ln(\sec(x))\quad\text{and}\quad\int\limits \tan(2x)\ dx=\frac12\ln\sec(2x)$$
yes -- that is correct

Answer 12

look at this

Answer 13

integrating factor I.F=\[e ^{\int\limits2\tan 2x dx} =e ^{\int\limits}\frac{ 2\sin 2x ~dx }{\cos 2x }=e ^{-\ln \cos 2x}=e ^{\ln \left( \cos 2x \right)^{-1}}=\frac{ 1 }{ \cos 2x }\]
complete solution is
\[y*\frac{ 1 }{\cos 2x }=\int\limits 3e ^{3x}\cos 2x*\frac{ 1 }{ \cos 2x }dx+c\]
\[y~\sec 2x=3\frac{ e ^{3x} }{3 }+c,y \sec 2x=e ^{3x}+c\]

Answer 14

@amrmagdy okay? what do you want me to notice? that is correct