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Ivahew [28]
3 years ago
10

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

Mathematics
1 answer:
Firdavs [7]3 years ago
6 0

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}

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Point B is a solution to the system of inequalities  

<h3>Further explanation  </h3>

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General formula  

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y = mx + c  

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The formula for a gradient (m) between 2 points    

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If the intersection of the x-axis (b, 0) and the y-axis (0, a) then the equation of the line:  

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It says inequality if there are symbol forms like <, >, ≤ or ≥  

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ax + by> c, ax + by ≥ c , ax + by <c , ax + by ≤ c  

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• a solid line because y includes equal to  

For line inequality (positive coefficient y)  

ax + by ≥ c then the solution is shaded upwards  

ax + by ≤ c then the solution is shaded down  

(Picture attached)

The line :

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intersect x-axis at point : 4,0

Intersect y-axis at point : 0,-2

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And the point located in the shaded plane of the two inequality is point B and I

<h3>Learn more</h3>

Linear inequality represented by the graph

brainly.com/question/9909671

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#LearnWithBrainly

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