What is the answer to solvefor d,d(r)=(sqrt(r))/pi-2r^2 ?

The answer to solvefor d,d(r)=(sqrt(r))/pi-2r^2 is d=(sqrt(r)-2pir^2)/(pir);\quad r\ne 0

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solvefor d,d(r)=(sqrt(r))/pi-2r^2

solve for

d, d (r) = \frac{r}{π} - 2 r^{2}

d = \frac{r - 2 π r ^{2}}{π r}; r \neq = 0

Solution steps

d (r) = \frac{r}{π} - 2 r^{2}

Divide both sides by

r; r \neq = 0

d = \frac{r - 2 π r ^{2}}{π r}; r \neq = 0

y^{^{''}} + y = - e^{3} + e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{2}

\frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} + 64 y = (x + 1) e^{x}

y^{^{''}} - 5 y^{^{'}} + 2 y = x e^{x}

y^{^{''}} + 3 y^{^{'}} - 4 y = 3 e^{x}

What is the answer to solvefor d,d(r)=(sqrt(r))/pi-2r^2 ?
The answer to solvefor d,d(r)=(sqrt(r))/pi-2r^2 is d=(sqrt(r)-2pir^2)/(pir);\quad r\ne 0