The solution of differential equation {d^{4} y}{d x^{4}} + 4 y = 0 is

Correct Answer is : y=e−x(c1cos⁡2x+c2sin⁡2x)+ex(c3cos⁡2x+c4sin⁡2x)y = e^{-x} (c₁ 2x + c₂ 2x) + e^{x} (c₃ 2x + c₄ 2x)y=e−x(c1cos2x+c2sin2x)+ex(c3cos2x+c4sin2x)

y = e^{-2 x} (c₁ x + c₂ x) + e² x} (c₃ x + c₄ x)

y = e^{-x} (c₁ x + c₂ x) + e^{x} (c₃ x + c₄ x)

y = e^{-x} (c₁ 2x + c₂ 2x) + e^{x} (c₃ 2x + c₄ 2x)

y = e^{-2 x} (c₁ x + c₂ x) + e² x} (c₃ x + c₄ x)

The solution of differential equation {d^{4} y}{d x^{4}} + 4 y = 0 is

Correct Answer is : y=e−x(c1cos⁡2x+c2sin⁡2x)+ex(c3cos⁡2x+c4sin⁡2x)y = e^{-x} (c₁ 2x + c₂ 2x) + e^{x} (c₃ 2x + c₄ 2x)y=e−x(c1cos2x+c2sin2x)+ex(c3cos2x+c4sin2x)

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HCU MCA 2013 Paper

Question : 64 of 75

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The solution of differential equation

\frac{d^{4} y}{d x^{4}} + 4 y = 0

$y = e^{-2 x} (c_{1} \cos x + c_{2} \sin x) + e^{2 x} (c_{3} \cos x + c_{4} \sin x)$
$y = e^{-x} (c_{1} \cos x + c_{2} \sin x) + e^{x} (c_{3} \cos x + c_{4} \sin x)$
$y = e^{-x} (c_{1} \cos 2x + c_{2} \sin 2x) + e^{x} (c_{3} \cos 2x + c_{4} \sin 2x)$
$y = e^{-2 x} (c_{1} \cos x + c_{2} \sin x) + e^{2 x} (c_{3} \cos x + c_{4} \sin x)$

Validate

Solution:

m^{4} + 4 = 0

\Rightarrow m^{4} + 4 + 4 m^{2} - 4 m^{2} = 0

\Rightarrow (m^{2} + 2)^{2} - (2 m)^{2} = 0

\Rightarrow (m^{2} + 2 + 2 m)(m^{2} + 2 - 2 m) = 0

Now;

(m^{2} + 2 + 2 m) = 0 \text{ and } (m^{2} + 2 - 2 m) = 0

m = -\frac{4 \pm \sqrt{4 - 4 \times 2}}{2}

and

m = \frac{4 \pm \sqrt{4 - 4 \times 2}}{2}

= \frac{-4 \pm \sqrt{4 i^{2}}}{2}

= \frac{-4 \pm \sqrt{4 i^{2}}}{2}

and

\;\;\;\;\;\;\;= { 4 ±√ { 4 i ^ { 2 } } } /{ 2 }

= \frac{4 \pm \sqrt{4 i^{2}}}{2}

= \frac{-4 \pm 2 i}{2}

and

\;\;\;\;\;\;\;\;\;\;= { 4 ±{ 2 i } } /{ 2 }

= \frac{4 \pm 2 i}{2}

= -2 \pm i

and

\;\;\;\;\;\;\;\;\;\;\;\;\;= { 2 ±{ i } }

∴

= 2 \pm i

∴

m = 2 \pm i,\, -2 \pm i

CF = e^{-2 x} (c_{1} \cos x + c_{2} \sin x) + e^{2 x} (c_{3} \cos x + c_{4} \sin x)

∴Complete solution, y = CF + PI

PI = 0

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