What is the solution for x^2y^'+y= x/y ,y(1)=4 ?

The solution for x^2y^'+y= x/y ,y(1)=4 is y=sqrt(-2e^{2/x)\Ei(-2/x)+(16+2e^2\Ei(-2))e^{(2-2x)/x}}

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

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

y = - 2 e^{\frac{2}{x}} Ei (- \frac{2}{x}) + (16 + 2 e^{2} Ei (- 2)) e^{\frac{2 - 2 x}{x}}

Solution steps

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

Solve Bernoulli ODE:

y = - 2 e^{\frac{2}{x}} Ei (- \frac{2}{x}) + (16 + 2 e^{2} Ei (- 2)) e^{\frac{2 - 2 x}{x}}

y = - 2 e^{\frac{2}{x}} Ei (- \frac{2}{x}) + (16 + 2 e^{2} Ei (- 2)) e^{\frac{2 - 2 x}{x}}

f (x) = x + 2, - 2, 1

\frac{( y ^{^{'}} - e - t + 4 )}{y} = - 4

x \to 0 lim (\frac{7 ^{s i n (x)} - 1}{x})

\int x^{2} \cdot cos^{3} (7) x^{3} d x

\frac{d y}{d x} + x^{2} sec (y) = 0

What is the solution for x^2y^'+y= x/y ,y(1)=4 ?
The solution for x^2y^'+y= x/y ,y(1)=4 is y=sqrt(-2e^{2/x)\Ei(-2/x)+(16+2e^2\Ei(-2))e^{(2-2x)/x}}