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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 66 of 160

Marks: +1, -0

If

y=e^{ax}(\cos bx+\sin bx)

satisfies the equation

\frac{d^2y}{dx^2}-K\frac{dy}{dx}+Ly=0

L+bK=

0
$(a+b)^2$
$a^2-b^2$
$a^2+b^2$

Solution:

We have,

y=e^{a x}(\cos b x+\sin b x)

\;\frac{dy}{dx}=e^{a x}(-\sin b x \cdot b+\cos b x \cdot b)

+a \cdot e^{a x}(\cos b x+\sin b x)

\;\frac{dy}{dx}=b e^{a x}(\cos b x-\sin b x)+a y

\Rightarrow \;\; \frac{d^{2}y}{dx^{2}}=b[e^{a x}(-\sin b x \cdot b-\cos b x \cdot b)

+a \cdot e^{a x}(\cos b x-\sin b x)]+a \;\frac{dy}{dx}

\Rightarrow \;\; \frac{d^{2}y}{dx^{2}}=b

[-b e^{a x}(\sin b x+\cos b x)+a e^{a x}(\cos b x-\sin b x)]

+a \;\frac{dy}{dx}=0

\Rightarrow \;\; \frac{d^{2}y}{dx^{2}}=b[(-b)(y)+a e^{a x}(\cos b x-\sin b x)]+a \;\frac{dy}{dx}

\Rightarrow \;\; \frac{d^{2}y}{dx^{2}}-a \;\frac{dy}{dx}+b^{2}y-a b e^{a x}(\cos b x-\sin b x)=0

\Rightarrow \;\; \frac{d^{2}y}{dx^{2}}-a \;\frac{dy}{dx}+b^{2}y-a\left(\;\frac{dy}{dx}-a y\right)=0

\Rightarrow \;\; \frac{d^{2}y}{dx^{2}}-2a \;\frac{dy}{dx}+(b^{2}+a^{2})y=0

On comparing with

\;\frac{d^{2}y}{dx^{2}}-k \;\frac{dy}{dx}+L y=0

\therefore \;\; K=2a,\: L=b^{2}+a^{2}

\therefore \;\; L+b k=b^{2}+a^{2}+2ab=(a+b)^{2}

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