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carrottopsadly

01.07.2019 •

Mathematics

Find f. f ''(x) = 5 + cos(x), f(0) = −1, f(3π/2) = 0

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Ответ:

tonya3498

11.03.2022

The required function is $f(x)=\dfrac{5}{2}x^2-\cos x-\dfrac{15\pi}{4}x$ .

Given:

The given differential equation is:

$f''(x)=5+\cos (x)$

The initial values are $f(0)=-1,f\left(\dfrac{3\pi}{2}\right)=0$ .

To find:

The function .

Explanation:

We have,

$f''(x)=5+\cos (x)$

Integrate both sides with respect to .

$\int f''(x)dx=\int (5+\cos (x))dx$

$f'(x)=5x+\sin (x)+C_1$

Integrate both sides with respect to .

$\int f'(x)dx=\int(5x+\sin (x)+C_1)dx$

$f(x)=\dfrac{5}{2}x^2-\cos (x)+C_1x+C_2$ ...(i)

We have f(0)=-1 . So,

$f(0)=\dfrac{5}{2}(0)^2-\cos (0)+C_1(0)+C_2$

-1=0-1+0+C_2

0=C_2

We have $f\left(\dfrac{3\pi}{2}\right)=0,C_2=0$ . So,

$f\left(\dfrac{3\pi}{2}\right)=\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)^2-\cos \left(\dfrac{3\pi}{2}\right)+C_1\left(\dfrac{3\pi}{2}\right)+0$

$0=\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)^2-0+C_1\left(\dfrac{3\pi}{2}\right)+0$

$-\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)^2=C_1\left(\dfrac{3\pi}{2}\right)$

On further simplification, we get

$-\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)^2\times \dfrac{2}{3\pi}=C_1$

$-\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)=C_1$

$-\dfrac{15\pi}{4}=C_1$

Substituting $C_1=-\dfrac{15\pi}{4}\text{ and }C_2=0$ , we get

$f(x)=\dfrac{5}{2}x^2-\cos x-\dfrac{15\pi}{4}x$

Therefore, the required function is $f(x)=\dfrac{5}{2}x^2-\cos x-\dfrac{15\pi}{4}x$ .

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Ответ:

emily12403

11.03.2022

Integrate both sides twice to recover the general form of f(x) :

$f''(x)=5+\cos x\implies f'(x)=5x+\sin x+C_1\implies f(x)=\dfrac52x^2-\cos x+C_1x+C_2$

With the given boundary conditions, we have

$\begin{cases}-1+C_2=-1\\\\\dfrac52\left(\dfrac{3\pi}2\right)^2+\dfrac{3\pi}2C_1+C_2=0\end{cases}\implies C_1=-\dfrac{15\pi}4,C_2=0$

So,

$f(x)=\dfrac52x^2-\cos x-\dfrac{15\pi}4x$

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Ответ:

sanafarghal

25.05.2022

a cubic function

Step-by-step explanation:

because it is has a domain and range of all real numbers

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