What is the solution for (y^2+4)dx-1/((2x+5)^2)dy=0 ?

The solution for (y^2+4)dx-1/((2x+5)^2)dy=0 is y=2tan((8x^3)/3+20x^2+50x-c_{1})

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(y^2+4)dx-1/((2x+5)^2)dy=0

(y^{2} + 4) d x - \frac{1}{( 2 x + 5 ) ^{2}} d y = 0

y = 2 tan (\frac{8 x ^{3}}{3} + 20 x^{2} + 50 x - c_{1})

Solution steps

(y^{2} + 4) d x - \frac{1}{( 2 x + 5 ) ^{2}} d y = 0

Solve separable ODE:

y = 2 tan (\frac{8 x ^{3}}{3} + 20 x^{2} + 50 x - c_{1})

y = 2 tan (\frac{8 x ^{3}}{3} + 20 x^{2} + 50 x - c_{1})

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\int_{- 8}^{8} \frac{1}{2} (64 - x^{2}) d x

n = 1 \sum \infty \frac{8}{n ^{4}}

y^{^{'}} sin (x) + y cos (x) = 0

What is the solution for (y^2+4)dx-1/((2x+5)^2)dy=0 ?
The solution for (y^2+4)dx-1/((2x+5)^2)dy=0 is y=2tan((8x^3)/3+20x^2+50x-c_{1})