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Popular >

y^'x+4y=1

Solution

y^{^{'}} x + 4 y = 1

Solution

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

Solution steps

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

Solve separable ODE:

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

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

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\frac{d y}{d x} = 90 y x^{9}

Frequently Asked Questions (FAQ)

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