What is the solution for yy^'=4x+9 ?

The solution for yy^'=4x+9 is y=sqrt(4x^2+18x+c_{1)},y=-sqrt(4x^2+18x+c_{1)}

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yy^'=4x+9

y y^{^{'}} = 4 x + 9

y = 4 x^{2} + 18 x + c_{1}, y = - 4 x^{2} + 18 x + c_{1}

Solution steps

y y^{'} (x) = 4 x + 9

Solve separable ODE:

y = 4 x^{2} + 18 x + c_{1}, y = - 4 x^{2} + 18 x + c_{1}

y = 4 x^{2} + 18 x + c_{1}, y = - 4 x^{2} + 18 x + c_{1}

\frac{df}{d t} f (t) = (sin (t^{2}) + 1)^{4}

\frac{d y}{d x} = \frac{15 x ^{2}}{y}

\frac{d y}{d x} = \frac{( x + 2 )}{2 y}, y (0) = 1

a y^{^{'}} + b y = c

\frac{d x}{d t} = 18 - \frac{9 x}{10}

What is the solution for yy^'=4x+9 ?
The solution for yy^'=4x+9 is y=sqrt(4x^2+18x+c_{1)},y=-sqrt(4x^2+18x+c_{1)}