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Popular >

sin(x)y^'+y=0

Solution

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

Solution

y = \frac{c _{1}}{tan ( \frac{x}{2} )}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{tan ( \frac{x}{2} )}

y = \frac{c _{1}}{tan ( \frac{x}{2} )}

Popular Examples

2y+3+(2x-3)(dy)/(dx)=0

2 y + 3 + (2 x - 3) \frac{d y}{d x} = 0

(4+x^2)(dy)/(dx)+2xy=4x

(4 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x

y^'-3/4 y=x^4(y^1)/3

y^{^{'}} - \frac{3}{4} y = x^{4} \frac{y ^{1}}{3}

y^'=((ln^4(x)))/((x*y)^2),y(1)=2

y^{^{'}} = \frac{( ln ^{4} ( x ) )}{( x \cdot y ) ^{2}}, y (1) = 2

y^{^{'}} + 16 x y = 0

Frequently Asked Questions (FAQ)

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