Question

Resolva a equação diferencial linear:
dy/dx+y=e^3x

Answer 1

$\frac{dy}{dx}+y= e^{3x}$ ou $\frac{dy}{dx+y} = e^{3x} ?$

$dy + y. dx = e^{3x} .dx$

$\int\limits^._. {} \, dy + \int\limits^._. {y} \, dx = \int\limits^._. { e^{3x} } \, dx$

$y+yx = \frac{1}{3} e^{3x}$

$y(1+x) = \frac{1}{3} e^{3x}$

$y = \frac{ e^{3x} }{3(1+x)} =\frac{ e^{3x} }{3+3x}$

Answer 2

$\sf{p(x)=1}\\\sf{\mu(x)=e^{\int p(x)~dx}}\\\sf{\mu(x)=e^{\int~dx}\implies \mu(x)=e^x}\\\sf{\mu(x)\dfrac{dy}{dx}] +\mu(x)~y=\mu(x)\cdot e^{3x}}\\\sf{e^x\dfrac{dy}{dx}+e^x\cdot y=e^x\cdot e^{3x}}\\\sf{\dfrac{d}{dx}(y\cdot~e^x)=e^{4x}}\\\sf{d(y\cdot e^x)=e^{4x}~dx}\\\displaystyle\sf{\int d(y\cdot e^x)=\int e^{4x}~dx}\\\sf{y\cdot e^x=\dfrac{1}{4}e^{4x}+k}\\\sf{y(x)=\dfrac{1}{4}\cdot\dfrac{e^{4x}}{e^x}+\dfrac{k}{e^x}}\\\sf{y(x)=\dfrac{1}{4}e^{3x}+\dfrac{k}{e^x}}$