If the solution of dy/dx = {y³ {x}}{{x} e¹/y²}, y(0)=1 is 1/y² = _{e}(f(x)), then f(x)=

Correct Answer is : e−4sin⁡xe - 4 {x}e−4sinx

Correct Answer is : e−4sin⁡xe - 4 {x}e−4sinx

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AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 78 of 160

Marks: +1, -0

If the solution of

\frac{dy}{dx} = \frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1/y^2}}, y(0)=1

\frac{1}{y^2} = \log_{e}(f(x))

f(x)=

Solution: 👈: Video Solution

78. (d)

\frac{dy}{dx} = \frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1/y^2}}

\Rightarrow \int \frac{e^{1/y^2}}{y^3} dy = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx

On putting

\frac{1}{y^2} = t \Rightarrow -\frac{2}{y^3} dy = dt

and on putting

u = \sqrt{x} \Rightarrow du = \frac{dx}{2 \sqrt{x}}

we get

\int \frac{e^t dt}{-2} = \int 2 \cos u du

\Rightarrow -\frac{1}{2} e^t = 2 \sin u + k

\Rightarrow -\frac{1}{2} e^{1/y^2} = 2 \sin \sqrt{x} + k

\because y(0)=1

means at

x=0, y=1

\Rightarrow -\frac{1}{2} e^{1/y} = 2 \sin 0 + k

\Rightarrow k = -\frac{e}{2}

\therefore -\frac{1}{2} e^{1/y^2} = 2 \sin \sqrt{x} - \frac{e}{2}

Taking log both sides with base

e

\log \left( -\frac{1}{2} e^{1/y^2} \right) = \log \left[ (2 \sin \sqrt{x}) - \frac{e}{2} \right]

\Rightarrow \log \left( -\frac{1}{2} \right) + \log \left( e^{1/y^2} \right) = \log \left( 2 \sin \sqrt{x} - \frac{e}{2} \right)

\Rightarrow \log 1 - \log(-2) + \frac{1}{y^2} = \log \left( 2 \sin \sqrt{x} - \frac{e}{2} \right)

\Rightarrow \frac{1}{y^2} = \log \left[ \left( 2 \sin \sqrt{x} - \frac{e}{2} \right) \times (-2) \right]

\Rightarrow \frac{1}{y^2} = \log (e - 4 \sin \sqrt{x})

\therefore f(x) = e - 4 \sin \sqrt{x}

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