Question

Solve the differential equation: `dy/dx - y = 2x`

Solution:

The differential equation `dy/dx - y = 2x` is in the form of `dy/dx + py = Q`, where P = −1 and Q = 2x

∴ I.F. = `e^(intP dx) = square`

∴ The solution of the linear differential equation is

∴ `y square = int 2x square dx + c`

= `2{x int e^(-x) dx - int e^(-x) dx * d/dx (x) dx} + c`

= `2{x int square/square - int square/square dx} + c`

∴ `ye^(−x) = −2xe^(−x) + 2∫e^(−x) dx + c`

∴ `ye^(-x) = -2 xe^(-x) + 2((e^(-x))/-1) + c`

∴ `y + square = ce^x` is the required solution of the given differential equation. 

Answer 1

The differential equation `dy/dx - y = 2x` is in the form of `dy/dx + py = Q`, where P = −1 and Q = 2x

\[\text{I.F.} = \text{e}^{\int\text{P dx}} = \boxed{\text{e}^{-\text{x}}}\]

∴ The solution of the linear differential equation is

∴ \[\text{y}\boxed{\text{e}^{-\text{x}}} = \int 2\text{x}\boxed{\text{e}^{-\text{x}}} \text{dx + c}\]

= `2{x int e^(-x) dx - int e^(-x) dx * d/dx (x) dx} + c`

= \[2\left\{x\int\frac{\boxed{e^{-x}}}{\boxed{-1}}-\int\frac{\boxed{e^{-x}}}{\boxed{-1}}dx\right\}+c\]

∴ `ye^(−x) = −2xe^(−x) + 2∫e^(−x) dx + c`

∴ `ye^(-x) = -2 xe^(-x) + 2((e^(-x))/-1) + c`

∴ y + \[\boxed{2\text{x} + 2}\] = ce^x is the required solution of the given differential equation.