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3 years ago

PLEASE HELPPPPP ME PLEASE I DONT KNOW WHAT TO DO ☹️#5

Mathematics

1 answer:

kow [346]3 years ago

3 0

Answer:

I can't see your question

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Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$

The general solution is

$y=y_{h}+y_{p}$

from (2) ad (3)

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

6 0

3 years ago

5 ft

Effectus [21]

the answer is 3 ft because since the two triangles are diagonal to eachother (i forgot the word) they have the same side length ratio. the ratio of the bigger triangle is 10:6.

the other triangle has 1/2 of that ratio. 10/2=5 and 6/2=3

6 0

3 years ago

What is the standard form of <br> Y=2/5x+2

Romashka-Z-Leto [24]

The answer to your question is -10

7 0

3 years ago

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