Differential EquationsmediumFree

First-Order Differential Equation with Initial Value | JEE

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

Question

The solution of the differential equation

\dfrac{dx}{dy} - \dfrac{x\log x}{1+\log x} = \dfrac{e^y}{1+\log x}

, if

y(1) = 0

, is:

x^x = e^{ye^y}

correct

e^y = x^{e^y}

x^x = ye^y

DNone of these

Solution

Step 1: Multiply through by

(1 + \log x)

(1+\log x)\frac{dx}{dy} - x\log x = e^y.

Step 2: Multiply by

e^{-y}

and notice the left side is an exact derivative, since

d(x\log x) = (1+\log x)\,dx

e^{-y}(1+\log x)\,dx - e^{-y}x\log x\,dy = dy \Rightarrow d\big(e^{-y}\,x\log x\big) = dy.

Step 3: Integrate:

e^{-y}\,x\log x = y + c \Rightarrow x\log x = (y+c)e^{y}.

Step 4: Apply

y(1) = 0

(i.e.

x=1

y=0

1\cdot\log 1 = (0+c)e^0 \Rightarrow c = 0

. Hence

x\log x = ye^{y} \Rightarrow \log x^{x} = ye^{y} \Rightarrow x^{x} = e^{ye^{y}}.

Correct answer: (1)

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