What is the solution for y^'+(4y)/x =-3/x ,y(1)=-2 ?

The solution for y^'+(4y)/x =-3/x ,y(1)=-2 is y=-3/4-5/(4x^4)

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y^'+(4y)/x =-3/x ,y(1)=-2

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

y = - \frac{3}{4} - \frac{5}{4 x ^{4}}

Solution steps

y^{'} (x) + \frac{4 y}{x} = - \frac{3}{x}

Solve linear ODE:

y = - \frac{3}{4} - \frac{5}{4 x ^{4}}

y = - \frac{3}{4} - \frac{5}{4 x ^{4}}

\int - 2 cos (\frac{x}{2}) d x

x \to 2 lim (x^{2} + 3 x)

\int 9 x^{- 2} d x

\frac{d}{d x} (\frac{x - 2}{x + 2})

y = x e^{a x}

What is the solution for y^'+(4y)/x =-3/x ,y(1)=-2 ?
The solution for y^'+(4y)/x =-3/x ,y(1)=-2 is y=-3/4-5/(4x^4)