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If I=V/R and V=250 volts and R=50ohms,find the change in I resulting from an increase of 1 voltage in v and an increase of 0.50ohm in R. Using integrating factor |
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Answer» Step-by-step explanation: The first equation GIVEN, X-\frac{1}{x}=5x− x 1
=5 …….. (i) If we do the CUBE of equation (i) we get, \left(x-\frac{1}{x}\right)^{3}=5^{3}(x− x 1
) 3 =5 3
x^{3}-\left(\frac{1}{x}\right)^{3}-3 \TIMES x \times\left(\frac{1}{x}\right) \times\left(x-\frac{1}{x}\right)=125x 3 −( x 1
) 3 −3×x×( x 1
)×(x− x 1
)=125 x^{3}-\left(\frac{1}{x}\right)^{3}-3\times(x-\frac{1}{x})=125x 3 −( x 1
) 3 −3×(x− x 1
)=125 x^{3}-\left(\frac{1}{x}\right)^{3}-3 \times 5=125x 3 −( x 1
) 3 −3×5=125 Grouping the terms, x^{3}-\left(\frac{1}{x}\right)^{3}=125+15x 3 −( x 1
) 3 =125+15 x^{3}-\left(\frac{1}{x}\right)^{3}=140x 3 −( x 1
) 3 =140 |
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