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Popular >

y^'-7/5 y=0

Solution

y^{^{'}} - \frac{7}{5} y = 0

Solution

y = e^{\frac{7 t + c _{1}}{5}}

Solution steps

y^{'} (t) - \frac{7}{5} y = 0

Solve separable ODE:

y = e^{\frac{7 t + c _{1}}{5}}

y = e^{\frac{7 t + c _{1}}{5}}

Popular Examples

y^'+y=y(te^{t^2}+1)t,y(0)=1

y^{^{'}} + y = y (t e^{t^{2}} + 1) t, y (0) = 1

(dy)/(dx)y=(x^{x-3})

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

((x^3+1)/(x^2))(dy)/(dx)= 1/y ,y(0)=-1

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

e^{x+y}*(dy)/(dx)=x

e^{x + y} \cdot \frac{d y}{d x} = x

x^{^{'}} = x^{7}

Frequently Asked Questions (FAQ)

What is the solution for y^'-7/5 y=0 ?
The solution for y^'-7/5 y=0 is y=e^{(7t+c_{1})/5}