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Popular >

y^'=y^2+1,y(1)=0

Solution

y^{^{'}} = y^{2} + 1, y (1) = 0

Solution

y = tan (t - 1)

Solution steps

y^{'} (t) = y^{2} + 1

Solve separable ODE:

y = tan (t - 1)

y = tan (t - 1)

Graph

Popular Examples

d y = θ^{3} d θ

(dy)/(dx)y=((5x^2+2)(3x+9))/(8x-9)

\frac{d y}{d x} y = \frac{( 5 x ^{2} + 2 ) ( 3 x + 9 )}{8 x - 9}

(1-x^2)y^'=6xy,y(0)=3

(1 - x^{2}) y^{^{'}} = 6 x y, y (0) = 3

3e^{2x}(dy)/(dx)=-4*x/(y^2)

3 e^{2 x} \frac{d y}{d x} = - 4 \cdot \frac{x}{y ^{2}}

(dy)/(dx)=((1-y))/(xv)

\frac{d y}{d x} = \frac{( 1 - y )}{xv}

Frequently Asked Questions (FAQ)

What is the solution for y^'=y^2+1,y(1)=0 ?
The solution for y^'=y^2+1,y(1)=0 is y=tan(t-1)