Question

solve the differential equation to the indicated inintial conditions; $y+4y=0$ $y(0)=4$ $y(0)=6$

Answer 1

Here, question is incorrect as $y+4y=0$ is not a differential equation also $y(0)=4, y(0)=6$

is not possible as we are getting two different $y$ values at $x=0.$

Let us take the differential equation as $y'+4y=0$ and the boundary condition as $y(0)=4.$

$y'+4y=0 \\\Rightarrow \frac{dy}{dx}+4y=0\\ \Rightarrow \frac{dy}{dx}=-4y\\\Rightarrow \frac{dy}{y}=-4dx$

Integrating both sides, we get

$\int \frac{dy}{y}=\int(-4)dx \\\Rightarrow ln \ y = -4x+c \ ...(1)$

Now, using boundary condition as y(0)=4

$ln \ 4=-4\times0+c\Rightarrow c=ln \ 4$

From equation (1), we get

$ln \ y = -4x+ln \ 4 \\\Rightarrow 4x=ln\ 4 - ln\ y \\\Rightarrow 4x=ln\ (\frac{4}{y}) \\\Rightarrow e^{4x}=\frac{4}{y} \\\Rightarrow y.e^{4x}=4$