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Marat540 [252]
3 years ago
5

How many solutions does the equation −2y + 2y + 3 = 3 have? One Zero Infinitely many Three

Mathematics
2 answers:
AnnZ [28]3 years ago
3 0

Answer:

Infinitely many solutions.

Step-by-step explanation:

In the equation −2y + 2y + 3 = 3  we see only one variable, and that variable is of the first power.  Ordinarily, we'd say that this equation will have 1 solution.  However, if we combine like terms, we get 0 + 3 = 3, or 0 = 0, which is true for any and all y values.  Infinitely many solutions.

Vitek1552 [10]3 years ago
3 0

Answer:

Infinitely

Step-by-step explanation:

we have

-2y+2y+3=3

Group terms that contain the same variable

-2y+2y=3-3

Combine like terms

0=0 ----> the equation is true for any value of y

therefore

The equation has infinite solutions

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Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2
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The Laplace transform of the given initial-value problem

y' 5y = e^{4t}, y(0) = 2 is  mathematically given as

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is  mathematically given as

&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}

\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}

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\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}

In conclusion, Taking inverse Laplace tranoform

L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

Read more about Laplace tranoform

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