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Popular >

2y^'+y=0,y(0)=e-x/2

Solution

2 y^{^{'}} + y = 0, y (0) = e - \frac{x}{2}

Solution

y = e^{\frac{- x + c _{1}}{2}}

Solution steps

2 y^{'} (x) + y = 0

Solve separable ODE:

y = e^{\frac{- x + c _{1}}{2}}

y = e^{\frac{- x + c _{1}}{2}}

Popular Examples

(dy)/(dx)=(x^2-xy-x+y)/(xy-y^2)

\frac{d y}{d x} = \frac{x ^{2} - x y - x + y}{x y - y ^{2}}

(dy}{dx}=\frac{1+y^2)/x

\frac{d y}{d x} = \frac{1 + y ^{2}}{x}

x^2(y^2+1)dx+ysqrt(x^3+1)dy=0

x^{2} (y^{2} + 1) d x + y x^{3} + 1 d y = 0

sqrt(1-x^2)y^'=sqrt(1-y^2)

1 - x^{2} y^{^{'}} = 1 - y^{2}

y^'=y(1-y/(10))

y^{^{'}} = y (1 - \frac{y}{10})

Frequently Asked Questions (FAQ)

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