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Ordinary Differential Equations

If y=y(x) is ...

Question

If y=y(x) is the solution of the differential equation,xdy÷dx+2y=x^2 satisfying y(1)=1,they(1÷2) is equal to

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y = y (x)

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

y (1) = 1

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y = y (x)

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

y (1) = 1

y (\frac{1}{2})

y = y (x)

e^{x} \sqrt{1 - y^{2}} d x + (\frac{y}{x}) d y = 0,

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y = y (x)

\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = 0

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x \geq 0

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(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0,

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