What is the solution for y^'+(3y)/(100-x)= 4/5 ?

The solution for y^'+(3y)/(100-x)= 4/5 is y=-(2(x-100))/5+c_{1}(1000000-30000x+300x^2-x^3)

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y^'+(3y)/(100-x)= 4/5

y^{^{'}} + \frac{3 y}{100 - x} = \frac{4}{5}

y = - \frac{2 ( x - 100 )}{5} + c_{1} (1000000 - 30000 x + 300 x^{2} - x^{3})

Solution steps

y^{'} (x) + \frac{3 y}{100 - x} = \frac{4}{5}

Solve linear ODE:

y = - \frac{2 ( x - 100 )}{5} + c_{1} (1000000 - 30000 x + 300 x^{2} - x^{3})

y = - \frac{2 ( x - 100 )}{5} + c_{1} (1000000 - 30000 x + 300 x^{2} - x^{3})

\int \frac{tan ^{3} ( ln ( x ) )}{4 x} d x

\int t cos^{2} (t) d t

ln (2)

\frac{\partial}{\partial x} (x yz + x y^{3} + y z^{3} + z x^{3})

\frac{d x}{d t} = t x^{\frac{1}{3}}

What is the solution for y^'+(3y)/(100-x)= 4/5 ?
The solution for y^'+(3y)/(100-x)= 4/5 is y=-(2(x-100))/5+c_{1}(1000000-30000x+300x^2-x^3)