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Popular >

ty^'+y=1+y+y^2

Solution

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

Solution

y = tan (ln (t) + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (ln (t) + c_{1})

y = tan (ln (t) + c_{1})

Popular Examples

derivative of {f}(x+{g}(x))

\frac{d}{d x} (f (x) + g (x))

derivative of x^m

\frac{d}{d x} (x^{m})

y^{''}+8y^'+15y=0

y^{^{''}} + 8 y^{^{'}} + 15 y = 0

derivative of f(x)=(2x)/(9x^2+1)derivative of

f (x) = \frac{2 x}{9 x ^{2} + 1}

integral from 4 to 7 of x^2

\int_{4}^{7} x^{2} d x

Frequently Asked Questions (FAQ)

What is the solution for ty^'+y=1+y+y^2 ?
The solution for ty^'+y=1+y+y^2 is y=tan(ln(t)+c_{1})