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Popular >

y^'+1/3 y=0

Solution

y^{^{'}} + \frac{1}{3} y = 0

Solution

y = e^{\frac{- t - c _{1}}{3}}

Solution steps

y^{'} (t) + \frac{1}{3} y = 0

Solve separable ODE:

y = e^{\frac{- t - c _{1}}{3}}

y = e^{\frac{- t - c _{1}}{3}}

Popular Examples

y^'=sqrt(3)y,y(0)=14

y^{^{'}} = 3 y, y (0) = 14

(ln(y))^6(dy)/(dx)=x^6y

(ln (y))^{6} \frac{d y}{d x} = x^{6} y

y^{^{'}} = - 5 y - 10

3t^2(dx)/(dt)-9x+9=0

3 t^{2} \frac{d x}{d t} - 9 x + 9 = 0

(dy)/(dx)=(xsec(y))/(e^{x^2)}

\frac{d y}{d x} = \frac{x sec ( y )}{e ^{x^{2}}}

Frequently Asked Questions (FAQ)

What is the solution for y^'+1/3 y=0 ?
The solution for y^'+1/3 y=0 is y=e^{(-t-c_{1})/3}