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3 years ago

I=E/R + r solve for r

1 answer:

4 0

For like lower case r?

then,

I = E/R + r

multiple by R to the other side and subtract E

r= I * R - E

hope that helps

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3 years ago

Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution

Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

$y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\$

for y=e^{6x}:

$y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\$

Now we replace in the differential equation y'' − y' − 30y = 0

for y=e^{-5x}:

$25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0$

for y=e^{6x}:

$36e^{6x}-6e^{6x}-30=0\\36-6+30=0$

Now, to know if both function are linearly independent we calculate the Wronskian

$W(f,g)=fg'-f'g$

$W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0$

I hope this is useful for you

Best regard

7 0

3 years ago