Question

solve the separable DE
dx/dt -x^3=x

Answer 1

do you want to solve for x or find the derivative?

Answer 2

solve for x

Answer 3

well wait...its separable equations so we need to get dx/dt by itself and then take the integral of both sides. The right side calls for partial fractions which is what I am having trouble with, then we solve for x

Answer 4

the farthest I got is
\[\int\limits dt=\int\limits \frac{ 1 }{ x+x^3}\]

Answer 5

\[\begin{align*}\frac{dx}{dt}-x^3&=x\\\\
\frac{dx}{dt}&=x^3+x\\\\
\frac{dx}{x^3+x}&=dt\\\\
\int\frac{dx}{x^3+x}&=\int dt\\\\
\int\frac{dx}{x(x^2+1)}&=\int dt
\end{align*}\]
Partial fractions, or trig sub. I prefer the former:
\[\begin{align*}\frac{1}{x(x^2+1)}&=\frac{A}{x}+\frac{Bx+C}{x^2+1}\\\\
1&=A(x^2+1)+x(Bx+C)\\
1&=Ax^2+A+Bx^2+Cx\end{align*}~~\implies~~\begin{cases}A+B=0\\C=0\\A=1\end{cases}\]
which gives \(A=1,~B=-1,~C=0\).
\[\begin{align*}\int\frac{dx}{x(x^2+1)}&=\int dt\\\\
\int\left(\frac{1}{x}-\frac{1}{x^2+1}\right)~dx&=\int dt
\end{align*}\]

Answer 6

thank you!
can you go into detail on how you found A, B and C?

Answer 7

\[\begin{align*}\frac{1}{x(x^2+1)}&=\frac{A}{x}+\frac{Bx+C}{x^2+1}\\\\
\frac{1}{x(x^2+1)}&=\frac{A(x^2+1)}{x(x^2+1)}+\frac{x(Bx+C)}{x(x^2+1)}\\\\
\frac{1}{x(x^2+1)}&=\frac{A(x^2+1)}{x(x^2+1)}+\frac{x(Bx+C)}{x(x^2+1)}\\\\
1&=A(x^2+1)+x(Bx+C)\\\\
1&=Ax^2+A+Bx^2+Cx\\\\
1&=(A+B)x^2+Cx+A\end{align*}\]
In the last line, we group the like terms (the equivalent powers of \(x\)). From here, we match up the coefficients of the powers of \(x\) on both sides:
\[0x^2+0x+1=(A+B)x^2+Cx+A\]
which gives
\[\begin{cases}A+B=0\\C=0\\A=1\end{cases}\]