 # Find the solution of {f}{''}{\left({x}\right)}={8}{x}+ \sin{{x}} bobbie71G 2021-09-16 Answered

Find the solution of $f\left(x\right)=8x+\mathrm{sin}x$

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Solve $\frac{{d}^{2}f\left(x\right)}{{dx}^{2}}=8x+\mathrm{sin}\left(x\right):$
Integrate both sides with respect to x:
$\frac{df\left(x\right)}{dx}=\int \left(8x+\mathrm{sin}\left(x\right)\right)dx=4{x}^{2}-\mathrm{cos}\left(x\right)+{c}_{1}$, where ${c}_{1}$ is an arbitrary constant.
INTERMEDIATE STEPS:
Take the integral:
$\int \left(8x+\mathrm{sin}\left(x\right)\right)dx$
Integrate the sum term by term and factor out constants:
$=\int \mathrm{sin}\left(x\right)dx+8\int xdx$
The integral of $\mathrm{sin}\left(x\right)$ is $-\mathrm{cos}\left(x\right)$:
$=-\mathrm{cos}\left(x\right)+8\int xdx$
The integral of x is $\frac{{x}^{2}}{2}:$
$=4{x}^{2}-\mathrm{cos}\left(x\right)+\text{constant}$
Integrate both sides with respect to x:
$f\left(x\right)=\int \left(4{x}^{2}-\mathrm{cos}\left(x\right)+{c}_{1}\right)dx=\frac{4{x}^{3}}{3}-\mathrm{sin}\left(x\right)+x{c}_{1}+{c}_{2}$, where ${c}_{2}$ is an arbitrary constant.
INTERMEDIATE STEPS:
Take the integral:
$\int \left({c}_{1}+4{x}^{2}-\mathrm{cos}\left(x\right)\right)dx$
Integrate the sum term by term and factor out constants:
$={c}_{1}\int 1dx-\int \mathrm{cos}\left(x\right)dx+4\int {x}^{2}dx$
The integral of 1 is x:
$={c}_{1}x-\int \mathrm{cos}\left(x\right)dx+4\int {x}^{2}dx$
The integral of $\mathrm{cos}\left(x\right)$ is $\mathrm{sin}\left(x\right)$:
$=-\mathrm{sin}\left(x\right)+{c}_{1}x+4\int {x}^{2}dx$
The integral of ${x}^{2}$ is $\frac{{x}^{3}}{3}$:
$={c}_{1}x+\frac{4{x}^{3}}{3}-\mathrm{sin}\left(x\right)+\text{constant}$