Question

Solve the separable differential equation:dx/dt= x^2+ (1/9) and find the particular solution satisfying the initial condition: x(0)=6

Answer 1

Answer:

The particular solution satisfying the initial condition, $x(0)=6$, of the differential equation $\frac{dx}{dt}=x^2+\frac{1}{9}$ is $x=\frac{\tan \left(\frac{t+3\arctan \left(18\right)}{3}\right)}{3}$.

Step-by-step explanation:

A separable differential equation is any differential equation that we can write in the following form.

$N(y)\frac{dy}{dx}=M(x)$

We may find the solutions to certain separable differential equations by separating variables, integrating with respect to x, and ultimately solving the resulting algebraic equation for y.

To find the solution of the differential equation $\frac{dx}{dt}=x^2+\frac{1}{9}$ you must:

Separate the differential equation and integrate both sides.

$dx=(x^2+\frac{1}{9})\cdot dt\\ \\\frac{dx}{x^2+\frac{1}{9}} =dt\\\\\int {\frac{dx}{x^2+\frac{1}{9}}} =\int dt$

Solving $\int \frac{dx}{x^2+\frac{1}{9}}$

$\mathrm{Apply\:Integral\:Substitution:}\:x=\frac{1}{3}u\\\\\int \frac{3}{u^2+1}du\\\\3\cdot \int \frac{1}{u^2+1}du\\\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u^2+1}du=\arctan \left(u\right)\\\\3\arctan \left(u\right)\\\\\mathrm{Substitute\:back}\:u=\frac{x}{\frac{1}{3}}\\\\3\arctan \left(3x\right)\\\\\int \frac{1}{x^2+\frac{1}{9}}dx=3\arctan \left(3x\right)+C$

Therefore,

$\int {\frac{dx}{x^2+\frac{1}{9}}} =\int dt\\\\3\arctan \left(3x\right)+C=t+D\\\\3\arctan \left(3x\right)=t+D-C\\\\3\arctan \left(3x\right)=t+E$

Now, we use the initial condition $x(0)=6$ to find the value of the constant E.

$3\arctan \left(3(6)\right)=0+E\\E=3\arctan \left(18\right)$

Thus,

$3\arctan \left(3x\right)=t+3\arctan \left(18\right)$

and we solve for x,

$\frac{3\arctan \left(3x\right)}{3}=\frac{t}{3}+\frac{3\arctan \left(18\right)}{3}\\\\\arctan \left(3x\right)=\frac{t+3\arctan \left(18\right)}{3}\\\\\arctan \left(x\right)=a\quad \Rightarrow \quad \:x=\tan \left(a\right)\\\\3x=\tan \left(\frac{t+3\arctan \left(18\right)}{3}\right)\\\\x=\frac{\tan \left(\frac{t+3\arctan \left(18\right)}{3}\right)}{3}$