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Popular >

(1+x^2)y^'=-4xy

Solution

(1 + x^{2}) y^{^{'}} = - 4 x y

Solution

y = \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

Solution steps

(1 + x^{2}) y^{'} (x) = - 4 x y

Solve separable ODE:

y = \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

y = \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

Graph

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

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\frac{d x}{d t} = 18 - \frac{72 x}{100}

y^'+((10^3))/4 y-2*10^3=0

y^{^{'}} + \frac{( 1 0 ^{3} )}{4} y - 2 \cdot 1 0^{3} = 0

Frequently Asked Questions (FAQ)

What is the solution for (1+x^2)y^'=-4xy ?
The solution for (1+x^2)y^'=-4xy is y=(c_{1})/(1+2x^2+x^4)