What is the solution for (d^3y)/(dx^3)-cos(x)=0 ?

The solution for (d^3y)/(dx^3)-cos(x)=0 is y=-sin(x)+(c_{1}x^2)/2+c_{2}x+c_{3}

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(d^3y)/(dx^3)-cos(x)=0

\frac{d ^{3} y}{d x ^{3}} - cos (x) = 0

y = - sin (x) + \frac{c _{1} x ^{2}}{2} + c_{2} x + c_{3}

Solution steps

\frac{d ^{3} y}{d x ^{3}} - cos (x) = 0

Substitute

\frac{d ^{3} y}{d x ^{3}}

with

y^{'''} (x)

y^{^{'''}} (x) - cos (x) = 0

Isolate

y^{'''} (x) : y^{'''} (x) = cos (x)

Solve

y^{'''} (x) = cos (x) : y^{''} (x) = sin (x) + c_{1}

Solve

y^{''} (x) = sin (x) + c_{1} : y^{'} (x) = - cos (x) + c_{1} x + c_{2}

Solve

y^{'} (x) = - cos (x) + c_{1} x + c_{2} : y = - sin (x) + \frac{c _{1} x ^{2}}{2} + c_{2} x + c_{3}

y = - sin (x) + \frac{c _{1} x ^{2}}{2} + c_{2} x + c_{3}

\int_{6}^{8} \frac{28}{( x - 6 ) ^{3}} d x

\int r^{3} d r

\int \frac{2 x ^{2} + 5 x + 18}{x ( x ^{2} + 9 )} d x

(- 9, - 9), (- 6, 0)

7 \cdot 5^{x}

What is the solution for (d^3y)/(dx^3)-cos(x)=0 ?
The solution for (d^3y)/(dx^3)-cos(x)=0 is y=-sin(x)+(c_{1}x^2)/2+c_{2}x+c_{3}