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2 years ago

What is the value of term a^4?

Mathematics

1 answer:

klasskru [66]2 years ago

8 0

Answer:

Step-by-step explanation:

$\text{We have the recursive formula of a series:}\\\\a_1=5\\a_n=3a_{n-1}\\\\a_2=3a_{2-1}=3a_1\to a_2=3(5)=15\\\\a_3=3a_{3-2}=3a_2\to a_3=3(15)=45\\\\a_4=3a_{4-1}=3a_3\to a_4=3(45)=135$

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sveticcg [70]

Answer:

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Step-by-step explanation:

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2 years ago

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$