Question

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

Answer 1

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