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Solve the differential equation 4 dy/dx=((e^y)sin^2x)/(ysecx).
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I got the answer e^-y*(-4y-4) = (1/3)*(sin(x))^3 + C However, it does not appear to be correct.
\[4\frac{dy}{dx}= \frac{e^y sin^2x}{y secx}\] right?
yes
\[4\dfrac{y}{e^y} dy = sin^2x cos x dx\] take integral both sides.
\[-4\dfrac{y+1}{e^y}= \frac{sin^3x}{3} + C\]
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That's exactly the answer that I got.
so, why do you make question on your correct answer?
Did I enter something wrong?
yes, you did. (sin x)^3 not sin (x)^3
but I also tried sin^3(x) and it did not work.
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oh yes, you should put theta, not x
okay, got it thanks.
ok
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