# ODE Integral by partial fractions

• Apr 22nd 2013, 07:06 PM
mathlover10
ODE Integral by partial fractions
the solution technique is easy but I'm exhausted and having surgery...by the way what is the partial fraction expansion for this differential equation? $\displaystyle \LARGE \frac{dy}{dt}+\frac{ty}{1+t^2}=1-\frac{t^3y}{t^4+1}......e^{\int \frac{2t^{5}+t^{3}+t}{(t^2+1)(t^4+1)}}$ thanks very much!
• Apr 22nd 2013, 08:57 PM
mathguy25
Re: ODE Integral by partial fractions
$\displaystyle \frac{2t^5 + t^3 + t}{(t^2 + 1)(t^4 + 1)} = \frac{At + B}{t^2 + 1} + \frac{Ct^3 + Dt^2 + Et + F}{t^4 + 1}$

Multiply both sides by $\displaystyle (t^2 + 1)(t^4 + 1)$

$\displaystyle 2t^5 + t^3 + t = (At+B)(t^4+1) + (Ct^3 + Dt^2 + Et + F)(t^2 + 1)$

$\displaystyle 2t^5 + t^3 + t = At^5 + Bt^4 + At + B + Ct^5 + Dt^4 + Et^3 + Ft^2 + Ct^3 + Dt^2 + Et + F$

$\displaystyle 2t^5 + t^3 + t = (A + C)t^5 + (B + D)t^4 + (E + C)t^3 + (D + F)t^2 + (A + E)t + (B + F)$

Therefore, $\displaystyle A + C = 2, B + D = 0, E + C = 1, D + F = 0, A + E = 1, B + F = 0$

Then $\displaystyle B = -D$ and $\displaystyle F = -D$ so then $\displaystyle B = F$. Then $\displaystyle B + F = B + B = 0$ so $\displaystyle 2B = 0$. Then $\displaystyle B = 0$. Thus, $\displaystyle F = 0 = -0 = D$.

Note that $\displaystyle (A + C) - (E + C) = 2 - 1$. Then $\displaystyle A - E = 1$. Then $\displaystyle (A - E) + (A + E) = 1 + 1$ i.e. $\displaystyle 2A = 2$. Then $\displaystyle A = 1$. This implies that $\displaystyle C = 1$ and $\displaystyle E = 0$.

Thus, $\displaystyle A = C = 1$ and $\displaystyle B = D = E = F = 0$.

Thus, $\displaystyle \frac{2t^5 + t^3 + t}{(t^2+1)(t^4+1)} = \frac{t}{t^2+1} + \frac{t^3}{t^4 + 1}$

Now, apply direct u-substitute on each summand.
• Apr 22nd 2013, 09:28 PM
mathlover10
Re: ODE Integral by partial fractions
oh right, so for example in the situation that Q(x) has irreducible quadratic factors none of which is repeated...$\displaystyle \int \frac{Ax+B}{ax^2+bx+c}=\int\frac{dx}{x^2+a^2}= \frac{1}{a}tan^-1(\frac{x}{a})+C$
• Apr 23rd 2013, 02:34 AM
Prove It
Re: ODE Integral by partial fractions
Quote:

Originally Posted by mathlover10
the solution technique is easy but I'm exhausted and having surgery...by the way what is the partial fraction expansion for this differential equation? $\displaystyle \LARGE \frac{dy}{dt}+\frac{ty}{1+t^2}=1-\frac{t^3y}{t^4+1}......e^{\int \frac{2t^{5}+t^{3}+t}{(t^2+1)(t^4+1)}}$ thanks very much!

`\displaystyle \displaystyle \begin{align*} \frac{dy}{dt} + \left( \frac{t}{1 + t^2} \right) y &= 1 - \left( \frac{t^3}{t^4 + 1} \right) y \\ \frac{dy}{dt} + \left( \frac{t}{1 + t^2} + \frac{t^3}{t^4 + 1} \right) y &= 1 \end{align*}

Now evaluating the integrating factor:

$\displaystyle \displaystyle e^{\int{\frac{t}{1 + t^2} + \frac{t^3}{t^4 + 1}\,dt}} = e^{\frac{1}{2}\ln{\left( 1 + t^2 \right)} + \frac{1}{4} \ln{\left( 1 + t^4 \right)}} = e^{\ln{\left[ \left( 1 + t^2 \right) ^{\frac{1}{2}} \right]} + \ln{ \left[ \left( 1 + t^4 \right) ^{\frac{1}{4}} \right] }} = e^{\ln{\left[ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \right]}} = \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}}$

Multiplying both sides of the DE by the integrating factor gives

\displaystyle \displaystyle \begin{align*} \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \frac{dy}{dt} + \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \left( \frac{t}{1 + t^2} \right) y &= \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \\ \frac{d}{dt} \left[ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} y \right] &= \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \\ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} y &= \int{ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \,dt } \\ y &= \left( 1 + t^2 \right) ^2 \left( 1 + t^4 \right) ^4 \int{ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \,dt } \end{align*}

There is not an indefinite integral of this in terms of the elementary function, so that is as far as you can go.
• Apr 23rd 2013, 09:45 PM
mathlover10
Re: ODE Integral by partial fractions
thanks very much! so when you do the partial fraction expansion the 2nd term has 4 terms in the numerator because it is t^4? is it more correct to express it in this form with a constant added to the integral? \displaystyle \large \displaystyle \begin{align*} \frac{d}{dt} \left[ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} y \right] &= \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \\ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} y &+c= \int{ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \,dt } \\ y&= \left( 1 + t^2 \right) ^{-1/2} \left( 1 + t^4 \right) ^{-1/4} [c_{\alpha}+\int{ \left( 1 + t^2 \right) ^{\frac{1}{2}} \left( 1 + t^4 \right) ^{\frac{1}{4}} \,dt }] \end{align*} sometimes the coefficient system can also be represented by a matrix and solved by Gaussian elimination