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Bernoulli Differential Equation – Finding dy/dx | JEE

JEE Maths question with a full step-by-step solution.

Question

Suppose a solution of

(xy^3 + x^2y^7)\dfrac{dy}{dx} = 1

satisfies

y\left(\dfrac{1}{4}\right) = 1

. Then the value of

\dfrac{dy}{dx}

when

y = -1

is:

-\dfrac{3}{20}

-\dfrac{20}{3}

-\dfrac{5}{16}

-\dfrac{16}{5}

correct

Solution

Step 1: From the equation,

\dfrac{dy}{dx} = \dfrac{1}{xy^3 + x^2y^7}

, so we first need

x

when

y = -1

. Treat

x

as a function of

y

\frac{dx}{dy} = xy^3 + x^2y^7 \Rightarrow \frac{dx}{dy} - xy^3 = x^2y^7 \quad(\text{Bernoulli}).

Step 2: Divide by

x^2

and put

t = \dfrac{1}{x}

(so

\dfrac{dt}{dy} = -\dfrac{1}{x^2}\dfrac{dx}{dy}

\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 = \dfrac{y^4}{4}

\displaystyle\int y^7 e^{y^4/4}\,dy = e^{y^4/4}(y^4 - 4)

, so

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 = \dfrac{1}{4}

y = 1

, i.e.

t = 4

4 = 3 + Ce^{-1/4} \Rightarrow C = e^{1/4}

. Then at

y = -1

(

y^4 = 1

t = 4 - 1 + e^{1/4}e^{-1/4} = 3 + 1 = 4 \Rightarrow x = \frac{1}{4}.

Step 5: Substitute into

\dfrac{dy}{dx}

y = -1

x = \dfrac14

(so

y^3 = y^7 = -1

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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