What is the solution for (d^2y}{dx^2}=\frac{(6x^5))/5-3x ?

The solution for (d^2y}{dx^2}=\frac{(6x^5))/5-3x is y=(x^7}{35}-\frac{x^3)/2+c_{1}x+c_{2}

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(d^2y}{dx^2}=\frac{(6x^5))/5-3x

\frac{d ^{2} y}{d x ^{2}} = \frac{( 6 x ^{5} )}{5} - 3 x

y = \frac{x ^{7}}{35} - \frac{x ^{3}}{2} + c_{1} x + c_{2}

Solution steps

\frac{d ^{2} y}{d x ^{2}} = \frac{( 6 x ^{5} )}{5} - 3 x

Substitute

\frac{d ^{2} y}{d x ^{2}}

with

y^{''} (x)

y^{^{''}} (x) = \frac{( 6 x ^{5} )}{5} - 3 x

Solve

y^{''} (x) = \frac{6 x ^{5}}{5} - 3 x : y^{'} (x) = \frac{x ^{6}}{5} - \frac{3 x ^{2}}{2} + c_{1}

Solve

y^{'} (x) = \frac{x ^{6}}{5} - \frac{3 x ^{2}}{2} + c_{1} : y = \frac{x ^{7}}{35} - \frac{x ^{3}}{2} + c_{1} x + c_{2}

y = \frac{x ^{7}}{35} - \frac{x ^{3}}{2} + c_{1} x + c_{2}

n \to \infty lim (n 0.25 - \frac{n}{n})

y = \frac{x ^{2} + 1}{x + 1}

\int_{1}^{6} w w + 3 d w

y^{^{''}} + y^{^{'}} + 4 y = 5 x - 7

\int \frac{7}{5} sin (3 x) cos (\frac{x}{3}) d x

What is the solution for (d^2y}{dx^2}=\frac{(6x^5))/5-3x ?
The solution for (d^2y}{dx^2}=\frac{(6x^5))/5-3x is y=(x^7}{35}-\frac{x^3)/2+c_{1}x+c_{2}