What is the solution for y^'=sec^2(x)(1-3y),y(pi/4)=0 ?

The solution for y^'=sec^2(x)(1-3y),y(pi/4)=0 is y=-(e^{-3tan(x)+3})/3+1/3

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y^'=sec^2(x)(1-3y),y(pi/4)=0

y^{^{'}} = sec^{2} (x) (1 - 3 y), y (\frac{π}{4}) = 0

y = - \frac{e ^{- 3 t a n (x) + 3}}{3} + \frac{1}{3}

Solution steps

y^{'} (x) = sec^{2} (x) (1 - 3 y)

Solve separable ODE:

y = - \frac{e ^{- 3 t a n (x) + 3}}{3} + \frac{1}{3}

y = - \frac{e ^{- 3 t a n (x) + 3}}{3} + \frac{1}{3}

f (x) = x^{2} + 16, at x = 3

\int sin^{3} (5 x) cot (5 x) d x

6 y^{^{''}} + 9 y^{^{'}} = 0

\int \frac{1}{( x ^{2} - 2 x + 1 ) x ^{2} - 2 x} d x

\frac{\partial}{\partial z} (2 x^{2} y)

What is the solution for y^'=sec^2(x)(1-3y),y(pi/4)=0 ?
The solution for y^'=sec^2(x)(1-3y),y(pi/4)=0 is y=-(e^{-3tan(x)+3})/3+1/3