The particular integral of the differential equation d4 ydx4 - 4y - x4 will be at x - 2

Particular integral, P I = x^4D^4 + 4 = 14x^41 + D^44 = 14(1 D^44 )x^4 = 14(x^4 d^4 x^44 dx^4 ) = 14(x^4 4 × 3 × 2 × 14 ) = 14 (x^4 6) At x = 2, PI = 14 ...

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Standard XII

Mathematics

Degree of a Differential Equations

The particula...

Question

The particular integral of the differential equation $\frac{d^{4} y}{d x^{4}} + 4 y = x^{4}$ will be at x = 2

2.5

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Solution

The correct option is A 2.5
Particular integral, $P I = \frac{x^{4}}{D^{4} + 4} = \frac{1}{4} \frac{x^{4}}{1 + \frac{D^{4}}{4}} = \frac{1}{4} (1 - \frac{D^{4}}{4}) x^{4}$
$= \frac{1}{4} (x^{4} - \frac{d^{4} x^{4}}{4 d x^{4}}) = \frac{1}{4} (x^{4} - \frac{4 \times 3 \times 2 \times 1}{4}) = \frac{1}{4} (x^{4} - 6)$
At $x = 2, P I = \frac{1}{4} (2^{4} - 6) = \frac{10}{4} = 2.5$

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The particular integral of the differential equation d4ydx4+4y=x4 will be at x = 2 2.5

The correct option is A 2.5Particular integral, PI=x4D4+4=14x41+D44=14(1−D44)x4 =14(x4−d4x44dx4)=14(x4−4×3×2×14)=14(x4−6) At x=2,PI=14(24−6)=104=2.5

The particular integral of the differential equation $\frac{d^{4} y}{d x^{4}} + 4 y = x^{4}$ will be at x = 2

2.5