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Popular >

y^'=2^{x-y}

Solution

y^{^{'}} = 2^{x - y}

Solution

y = \frac{ln ( 2 ^{x} + c _{1} )}{ln ( 2 )}

Solution steps

y^{'} (x) = 2^{x - y}

Solve separable ODE:

y = \frac{ln ( 2 ^{x} + c _{1} )}{ln ( 2 )}

y = \frac{ln ( 2 ^{x} + c _{1} )}{ln ( 2 )}

Popular Examples

y^2y^'=2(t+ty^3)e^{t^2},y(0)=0

y^{2} y^{^{'}} = 2 (t + t y^{3}) e^{t^{2}}, y (0) = 0

y^'(x)=(2x-1)/(3y^2-3),y(1)=0

y^{^{'}} (x) = \frac{2 x - 1}{3 y ^{2} - 3}, y (1) = 0

\frac{d y}{d x} = \frac{2}{y}

(dy}{dx}=\frac{(x^2+2))/y

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

y^'=e^{-y}sin(x),y(pi/2)= 1/2

y^{^{'}} = e^{- y} sin (x), y (\frac{π}{2}) = \frac{1}{2}

Frequently Asked Questions (FAQ)

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