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Bernoulli Differential Equation – Finding dy/dx | JEE

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Question
Suppose a solution of (xy3+x2y7)dydx=1(xy^3 + x^2y^7)\dfrac{dy}{dx} = 1 satisfies y(14)=1y\left(\dfrac{1}{4}\right) = 1. Then the value of dydx\dfrac{dy}{dx} when y=−1y = -1 is:
A−320-\dfrac{3}{20}
B−203-\dfrac{20}{3}
C−516-\dfrac{5}{16}
D−165-\dfrac{16}{5}correct
Solution
Step 1: From the equation, dydx=1xy3+x2y7\dfrac{dy}{dx} = \dfrac{1}{xy^3 + x^2y^7}, so we first need xx when y=−1y = -1. Treat xx as a function of yy:
dxdy=xy3+x2y7⇒dxdy−xy3=x2y7(Bernoulli).\frac{dx}{dy} = xy^3 + x^2y^7 \Rightarrow \frac{dx}{dy} - xy^3 = x^2y^7 \quad(\text{Bernoulli}).
 Step 2: Divide by x2x^2 and put t=1xt = \dfrac{1}{x} (so dtdy=−1x2dxdy\dfrac{dt}{dy} = -\dfrac{1}{x^2}\dfrac{dx}{dy}):
dtdy+y3t=−y7,IF=e∫y3 dy=ey4/4.\frac{dt}{dy} + y^3 t = -y^7, \qquad \text{IF} = e^{\int y^3\,dy} = e^{y^4/4}.
 Step 3: Solve. With u=y44u = \dfrac{y^4}{4}, ∫y7ey4/4 dy=ey4/4(y4−4)\displaystyle\int y^7 e^{y^4/4}\,dy = e^{y^4/4}(y^4 - 4), so
t ey4/4=−ey4/4(y4−4)+C⇒t=1x=4−y4+Ce−y4/4.t\,e^{y^4/4} = -e^{y^4/4}(y^4 - 4) + C \Rightarrow t = \frac{1}{x} = 4 - y^4 + C e^{-y^4/4}.
 Step 4: Apply x=14x = \dfrac{1}{4} at y=1y = 1, i.e. t=4t = 4: 4=3+Ce−1/4⇒C=e1/44 = 3 + Ce^{-1/4} \Rightarrow C = e^{1/4}. Then at y=−1y = -1 (y4=1y^4 = 1):
t=4−1+e1/4e−1/4=3+1=4⇒x=14.t = 4 - 1 + e^{1/4}e^{-1/4} = 3 + 1 = 4 \Rightarrow x = \frac{1}{4}.
 Step 5: Substitute into dydx\dfrac{dy}{dx} at y=−1y = -1, x=14x = \dfrac14 (so y3=y7=−1y^3 = y^7 = -1):
xy3+x2y7=−14−116=−516⇒dydx=1−5/16=−165.xy^3 + x^2y^7 = -\frac{1}{4} - \frac{1}{16} = -\frac{5}{16} \Rightarrow \frac{dy}{dx} = \frac{1}{-5/16} = -\frac{16}{5}.
 Correct answer: (4)
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