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Popular >

y^'=-0.5y+x

Solution

y^{^{'}} = - 0.5 y + x

Solution

y = 2 x - 4 + \frac{c _{1}}{e ^{0.5 x}}

Solution steps

y^{'} (x) = - 0.5 y + x

Solve linear ODE:

y = 2 x - 4 + \frac{c _{1}}{e ^{0.5 x}}

y = 2 x - 4 + \frac{c _{1}}{e ^{0.5 x}}

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[1]

Frequently Asked Questions (FAQ)

What is the solution for y^'=-0.5y+x ?
The solution for y^'=-0.5y+x is y=2x-4+(c_{1})/(e^{0.5x)}