If a ≠ b , x ≠ nπ , n ∊ Z and y² = a²² x + b²² x , then d²y/dx² + y =

Correct Answer is : 1y(aby)21/y≤ft(ab/y)²y1(yab)2

Correct Answer is : 1y(aby)21/y≤ft(ab/y)²y1(yab)2

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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 67 of 160

Marks: +1, -0

y^2

a^2\cos^2 x + b^2\sin^2 x

\frac{d^2y}{dx^2}

+ y =

Solution:

It is given that,

a \neq b,\ x \neq n\pi,\ n \in z

and

y^{2}=a^{2}\cos^{2}x+b^{2}\sin^{2}x

Differentiate the above equation with respect to

x

2y\;\frac{dy}{dx}=-a^{2}\sin 2x+b^{2}\sin 2x

2y\;\frac{dy}{dx}=(b^{2}-a^{2})\sin 2x

Again differentiate the above equation with respect to

x

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

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

Multiply

y^{2}

on both sides,

y^{3}\;\frac{d^{2}y}{dx^{2}}+y^{2}\left(\frac{dy}{dx}\right)^{2}=y^{2}(b^{2}-a^{2})\cos 2x

y^{3}\;\frac{d^{2}y}{dx^{2}}=y^{2}(b^{2}-a^{2})\cos 2x-y^{2}\left(\frac{dy}{dx}\right)^{2}

Add

y^{4}

in the above equation.

y^{4}+y^{3}\;\frac{d^{2}y}{dx^{2}}=y^{4}+y^{2}(b^{2}-a^{2})\cos 2x-y^{2}\left(\frac{dy}{dx}\right)^{2}+y^{4}

Substitute the values in the above equation.

y^{4}+y^{3}\;\frac{d^{2}y}{dx^{2}}\;\;=y^{4}+y^{2}(b^{2}-a^{2})\cos 2x-y^{2}\left(\frac{dy}{dx}\right)^{2}+y^{4}

\;\; = (a^{2}\cos^{2}x+b^{2}\sin^{2}x)(b^{2}-a^{2})(\cos^{2}x-\sin^{2}x)-(b^{2}-a^{2}\sin x\cos x)^{2}+(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}

\;\; =a^{2}b^{2}\cos^{4}x+a^{2}b^{2}\sin^{2}x+2a^{2}b^{2}\sin^{2}x\cos^{2}x

\;\; =a^{2}b^{2}(\sin^{2}x+\cos^{2}x)^{2}

\;\; =a^{2}b^{2}

Or it can be written as

\;\frac{d^{2}y}{dx^{2}}+y=\frac{1}{y}\left(\frac{ab}{y}\right)^{2}

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