# Thread: ODE Integral by partial fractions

1. ## 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? $\LARGE \frac{dy}{dt}+\frac{ty}{1+t^2}=1-\frac{t^3y}{t^4+1}......e^{\int \frac{2t^{5}+t^{3}+t}{(t^2+1)(t^4+1)}}$ thanks very much!

2. ## Re: ODE Integral by partial fractions

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

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

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

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

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

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

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

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

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

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

Now, apply direct u-substitute on each summand.

3. ## 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... $\int \frac{Ax+B}{ax^2+bx+c}=\int\frac{dx}{x^2+a^2}= \frac{1}{a}tan^-1(\frac{x}{a})+C$

4. ## Re: ODE Integral by partial fractions

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? $\LARGE \frac{dy}{dt}+\frac{ty}{1+t^2}=1-\frac{t^3y}{t^4+1}......e^{\int \frac{2t^{5}+t^{3}+t}{(t^2+1)(t^4+1)}}$ thanks very much!
` \displaystyle \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:

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

Multiplying both sides of the DE by the integrating factor gives

\displaystyle \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.

5. ## 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? \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