Respuesta :
Integrate both sides twice to recover the general form of [tex]f(x)[/tex]:
[tex]f''(x)=5+\cos x\implies f'(x)=5x+\sin x+C_1\implies f(x)=\dfrac52x^2-\cos x+C_1x+C_2[/tex]
With the given boundary conditions, we have
[tex]\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[/tex]
So,
[tex]f(x)=\dfrac52x^2-\cos x-\dfrac{15\pi}4x[/tex]
The required function is [tex]f(x)=\dfrac{5}{2}x^2-\cos x-\dfrac{15\pi}{4}x[/tex].
Given:
The given differential equation is:
[tex]f''(x)=5+\cos (x)[/tex]
The initial values are [tex]f(0)=-1,f\left(\dfrac{3\pi}{2}\right)=0[/tex].
To find:
The function [tex]f[/tex].
Explanation:
We have,
[tex]f''(x)=5+\cos (x)[/tex]
Integrate both sides with respect to [tex]x[/tex].
[tex]\int f''(x)dx=\int (5+\cos (x))dx[/tex]
[tex]f'(x)=5x+\sin (x)+C_1[/tex]
Integrate both sides with respect to [tex]x[/tex].
[tex]\int f'(x)dx=\int(5x+\sin (x)+C_1)dx[/tex]
[tex]f(x)=\dfrac{5}{2}x^2-\cos (x)+C_1x+C_2[/tex] ...(i)
We have [tex]f(0)=-1[/tex]. So,
[tex]f(0)=\dfrac{5}{2}(0)^2-\cos (0)+C_1(0)+C_2[/tex]
[tex]-1=0-1+0+C_2[/tex]
[tex]0=C_2[/tex]
We have [tex]f\left(\dfrac{3\pi}{2}\right)=0,C_2=0[/tex]. So,
[tex]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[/tex]
[tex]0=\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)^2-0+C_1\left(\dfrac{3\pi}{2}\right)+0[/tex]
[tex]-\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)^2=C_1\left(\dfrac{3\pi}{2}\right)[/tex]
On further simplification, we get
[tex]-\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)^2\times \dfrac{2}{3\pi}=C_1[/tex]
[tex]-\dfrac{5}{2}\left(\dfrac{3\pi}{2}\right)=C_1[/tex]
[tex]-\dfrac{15\pi}{4}=C_1[/tex]
Substituting [tex]C_1=-\dfrac{15\pi}{4}\text{ and }C_2=0[/tex], we get
[tex]f(x)=\dfrac{5}{2}x^2-\cos x-\dfrac{15\pi}{4}x[/tex]
Therefore, the required function is [tex]f(x)=\dfrac{5}{2}x^2-\cos x-\dfrac{15\pi}{4}x[/tex].
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