Question

Solve dy\dx + xsin(2y) = x^3cos^2y

Answer 1

\[\begin{align*}
\frac{dy}{dx}+x\sin2y&=x^3\cos^2y\\
(x\sin2y-x^3\cos^2y)~dx+dy&=0
\end{align*}\]
Uh... any takers?

Answer 2

\[M_y=2x\cos2y+2x^3\cos y\sin y=2x\cos2y+x^3\sin2y\\
N_x=0\]
Hmm...

Answer 3

Integrating factor(s):
\[\mu(x)=\exp\left(\int\frac{M_y-N_x}{N}~dx\right)\\
\mu(x)=\exp\left(\int (2x\cos2y+x^3\sin2y)~dx\right)\]
\[\eta(y)=\exp\left(\int\frac{N_x-M_y}{M}~dy\right)\\
\eta(y)=\exp\left(\int\frac{2x\cos2y+x^3\sin2y}{x^3\cos^2y-x\sin2y}~dy\right)\]
\(\mu(x)\) would be easier to simplify but it's not in terms of \(x\) only...

The equation definitely isn't separable, since \(M_y\not=N_x\), which I showed in @ameramd's previous question is a required condition for separability.

Answer 4

My next guess would be some clever substitution, but not sure what it would be

Answer 5

thanx @SithsAndGiggles it's bernoulli Eqn.

Answer 6

@SithsAndGiggles

Answer 7

Giving you an energetic and warm Welcome to one and only 'Open Study'.

Answer 8

\[\frac{ dy }{ dx }+x \sin 2 y=x^3 \cos ^2y\]
divide by \[\cos ^2y\]
\[\sec ^2y \frac{ dy }{ dx }+x \frac{ 2\sin y \cos y }{ \cos ^2y }=x^3\]
\[\sec ^2y \frac{ dy }{ dx }+2x \tan y=x^3\]
put \[\tan y=t,\sec ^2y~\frac{ dy }{ dx }=\frac{ dt }{ dx }\]
\[\frac{ dt }{ dx }+2 x t=x^3\]
\[I.F=e ^{\int\limits 2x dx}=e ^{x^2}\]
solution is \[t e ^{x^2}=\int\limits x^3e ^{x^2}dx+c\]
\[t e ^{x^2}=\int\limits x^2 e ^{x^2}x dx+c=I1+c\]
put\[x^2=z,2xdx=dz,x dx=\frac{ dz }{ 2 }\]
\[I1=\frac{ 1 }{ 2 } \int\limits z e^z dz=\frac{ 1 }{ 2 }\left[ z e^z-\int\limits 1*e^z dz \right]=\frac{ 1 }{ 2}\left[ z e^z-e^z \right]\]
\[=\frac{ 1 }{ 2 }\left( z-1 \right)e^z=\frac{ 1 }{ 2 }\left( x^2-1 \right)e ^{x^2}\]
\[t e ^{x^2}=\frac{ 1 }{ 2 }\left( x^2-1 \right)e ^{x^2}+c\]
\[\tan y~ e ^{x^2}=\frac{ 1 }{ 2 }\left( x^2-1 \right)e ^{x^2}+c\]

Answer 9

y'sec^2y+2xtany=x3
u=tany
u'+2xu=x3
u=exp(-x2)[c+int(x3exp(x2)dx]=exp(-x2)[c+x2/2exp(x2)-1/2exp(x2)]
then
tany=cexp(-x2)+1/2(x2-1)