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== `sum_0^oo ` mathHandbook.com

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f(x) = x; `sum_(x=0)^oo`f(x) `int`y(x) dx `int y(x) (dx)^0.5` `int_0^1` sin(x) dx `d/dx`y(x) `(d^(1) y)/dx^(1)` y' `y^((1))(x)`








Input:
dsolve(y(1,x)^2-2*y^2-4*y-2)

Write:
`dsolve(y(1,x)^2-2*y^2-4*y-2)`



Compute: $$dsolve({(y^{(1)}(x))}^{2}-2\ {y}^{2}-4\ y-2)$$

Output: $$ dsolve({(y^{(1)}(x))}^{2}-2\ {y}^{2}-4\ y-2) == \frac {log(1+y)}{\sqrt {2}}=C_1+x\ and\ \frac {(-log(1+y))}{\sqrt {2}}=C_1+x $$ Result:$$\frac {log(1+y)}{\sqrt {2}}=C_1+x\ and\ \frac {(-log(1+y))}{\sqrt {2}}=C_1+x$$



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