Question

Solve the given initial-value problem. dy dx = x + 2y, y(0) = 7

Answer 1

`(\mathrm dy)/(\mathrm dx)=x+2y`

`(\mathrm dy)/(\mathrm dx)-2y=x`

`e^(-2x)(\mathrm dy)/(\mathrm dx)-2e^(-2x)y=xe^(-2x)`

`(\mathrm d)/(\mathrm dx)[e^(-2x)y]=xe^(-2x)`

`e^(-2x)y=\displaystyle\int xe^(-2x)\,\mathrm dx`

`e^(-2x)y=-\frac14e^(-2x)(1+2x)+C`

`y=-\frac{1+2x}4+Ce^(2x)`


`y(0)=7`

`7=-\frac{1+2\cdot0}4+Ce^(2\cdot0)`

`7=-\frac14+C`

`C=\frac{29}4`


`y=-\frac{1+2x}4+\frac{29}4e^(2x)`