What is the solution for (d^2y)/(dx^2)=sqrt(2x-3) ?

The solution for (d^2y)/(dx^2)=sqrt(2x-3) is y= 1/15 (2x-3)^{5/2}+c_{1}x+c_{2}

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(d^2y)/(dx^2)=sqrt(2x-3)

\frac{d ^{2} y}{d x ^{2}} = 2 x - 3

y = \frac{1}{15} (2 x - 3)^{\frac{5}{2}} + c_{1} x + c_{2}

Solution steps

\frac{d ^{2} y}{d x ^{2}} = 2 x - 3

Substitute

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

with

y^{''} (x)

y^{^{''}} (x) = 2 x - 3

Solve

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

Solve

y^{'} (x) = \frac{1}{3} (2 x - 3)^{\frac{3}{2}} + c_{1} : y = \frac{1}{15} (2 x - 3)^{\frac{5}{2}} + c_{1} x + c_{2}

y = \frac{1}{15} (2 x - 3)^{\frac{5}{2}} + c_{1} x + c_{2}

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What is the solution for (d^2y)/(dx^2)=sqrt(2x-3) ?
The solution for (d^2y)/(dx^2)=sqrt(2x-3) is y= 1/15 (2x-3)^{5/2}+c_{1}x+c_{2}