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Shtirlitz [24]
3 years ago
15

Can anyone help me with letter c please. f(t)

Mathematics
2 answers:
DochEvi [55]3 years ago
5 0

we have given that f(x)=2^x..

we need to find f(t)..

we know that we need to plug t in place of x in the equation f(x)=2^x.

f(t)=2^t .for example for f(4)=2^4

they have plugged 4 in place of x. so we plug in plug t in place of x .

almond37 [142]3 years ago
3 0

Well, they are just plugging the x in and evaluating the exponent. Here's what the answers would be:

f(1)=2^1=2\\ f(3)=2^3=8\\ f(t)=2^t

For the last one, it's no different then having an x there. t is just a variable as x is.

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Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

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We have the given differential equation: y′′+4y=5xcos(2x)

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y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

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y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

we get

y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

see attached

Step-by-step explanation:

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