If y=e^{x}( x) then x y₂+(x-1) y=

Correct Answer is : (2x−1)y1(2x-1) y₁(2x−1)y1

Correct Answer is : (2x−1)y1(2x-1) y₁(2x−1)y1

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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 66 of 160

Marks: +1, -0

y=e^{x}(\log x)

x y_{2}+(x-1) y=

Solution:

Consider the equation,

y=e^{x}(\log x)

Differentiate above w.r.t x ,

\frac{d y}{d x}= \frac{e^{x}}{x}+e^{x} \ln x

....(I)

Now,

\frac{d y}{d x}= \frac{e^{x}}{x}+y

Different above w.r.t x ,

x \frac{d^{2} y}{d x^{2}} = \frac{x^{2} e^{x} - x e^{x}}{x^{2}} + x \frac{d y}{d x}

x \frac{d^{2} y}{d x^{2}} + (x-1) y = \frac{x e^{x}(x-1)}{x^{2}}

+ x \frac{d y}{d x} + (x-1) y

x \frac{d^{2} y}{d x^{2}} + (x-1) y = \frac{e^{x}(x-1)}{x}

+ (x-1) y + x \frac{d y}{d x}

= e^{x} - \frac{e^{x}}{x} + (x-1) e^{x}

\ln x + x \frac{d y}{d x}

Further simplify the above,

x \frac{d^{2} y}{d x^{2}} + (x-1) y = (x-1) e^{x} \left\{ \frac{1}{x} + \ln x \right\}

+ x \frac{d y}{d x}

= (x-1) \left[ \frac{e^{x}}{x} + e^{x} \ln x \right] + x \frac{d y}{d x}

Consider equation (1),

x \frac{d^{2} y}{d x^{2}} + (x-1) y

= (x-1+x) \frac{d y}{d x}

x \frac{d^{2} y}{d x^{2}} + (x-1) y

= (2x-1) \frac{d y}{d x}

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