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Popular >

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

Solution

y^{^{'}} + \frac{y}{x} = \frac{1}{x ^{2}}

Solution

y = \frac{ln ( x )}{x} + \frac{c _{1}}{x}

Solution steps

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

Solve linear ODE:

y = \frac{ln ( x )}{x} + \frac{c _{1}}{x}

y = \frac{ln ( x )}{x} + \frac{c _{1}}{x}

Popular Examples

tangent of y=4sin(pix+y),(1,0)tangent of

y = 4 sin (π x + y), (1, 0)

sum from n=0 to infinity of 6(-2/3)^n

n = 0 \sum \infty 6 (- \frac{2}{3})^{n}

y^'=(y^4-3)/((4xy^3))

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

limit as x approaches-1 of 4x^2-3x+2

x \to - 1 lim (4 x^{2} - 3 x + 2)

integral of 1/(x^4sqrt(x^2-1))

\int \frac{1}{x ^{4} x ^{2} - 1} d x

Frequently Asked Questions (FAQ)

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