What is a solution to the differential equation $\frac{d y}{d x} = x + y$ $dy/dx=x+y$ ?

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Answer 1 · 2016-10-19T21:11:58

$y = C_{1} e^{x} - x - 1$ $y = C_1e^x-x-1$

Let $u = x + y$ $u = x + y$

$\Rightarrow \frac{d u}{d x} = \frac{d}{d x} (x + y) = 1 + \frac{d y}{d x}$ $=> (du)/dx = d/dx(x+y) = 1+dy/dx$

$\Rightarrow \frac{d y}{d x} = \frac{d u}{d x} - 1$ $=> dy/dx = (du)/dx-1$

Thus, making the substitutions into our original equation,

$\frac{d u}{d x} - 1 = u$ $(du)/dx-1 = u$

$\Rightarrow \frac{d u}{u + 1} = d x$ $=> (du)/(u+1) = dx$

$\Rightarrow \int \frac{d u}{u + 1} = \int d x$ $=> int(du)/(u+1)=intdx$

$\Rightarrow ln (u + 1) = x + C_{0}$ $=> ln(u+1) = x + C_0$

$\Rightarrow e^{ln (u + 1)} = e_{0}^{x + C_{0}}$ $=> e^(ln(u+1)) = e^(x+C_0)$

$\Rightarrow u + 1 = C_{1} e^{x}$ $=> u+1 = C_1e^x" "$ (where $C_{1} = e_{0}^{C_{0}}$ $C_1 = e^(C_0)$ )

Substituting $x + y = u$ $x+y = u$ back in,

$\Rightarrow x + y + 1 = C_{1} e^{x}$ $=> x + y + 1 = C_1e^x$

$:. y = C_1e^x-x-1$

What is a solution to the differential equation dy/dx=x+ydydx=x+ydy/dx=x+y?