What is the solution for y^'=1+9x-5y,y(1)=2 ?

The solution for y^'=1+9x-5y,y(1)=2 is y=(9x)/5-4/25+(9e^{5-5x})/(25)

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y^'=1+9x-5y,y(1)=2

y^{^{'}} = 1 + 9 x - 5 y, y (1) = 2

y = \frac{9 x}{5} - \frac{4}{25} + \frac{9 e ^{5 - 5 x}}{25}

Solution steps

y^{'} (x) = 1 + 9 x - 5 y

Solve linear ODE:

y = \frac{9 x}{5} - \frac{4}{25} + \frac{9 e ^{5 - 5 x}}{25}

y = \frac{9 x}{5} - \frac{4}{25} + \frac{9 e ^{5 - 5 x}}{25}

\frac{\partial}{\partial y} (ln (x^{2} + y^{2}))

\int \frac{9 x ^{2} + 5 x + 9}{( x ^{2} + 1 ) ^{2}} d x

ln (4 + x^{2})

\int_{1}^{2} 15 4 x^{2} - 3 d x

f (x) = x^{2} (1 - 6 x)

What is the solution for y^'=1+9x-5y,y(1)=2 ?
The solution for y^'=1+9x-5y,y(1)=2 is y=(9x)/5-4/25+(9e^{5-5x})/(25)