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$(1+t^2)y' +4ty =t, y(1) = 1/4 $

Given equation can be transformed into $(1+t^2)y' = (1 - 4y)t$ then

$\frac{dy}{dt} = \frac{t}{1+t^2}(1-4y)$ then

$\frac{1}{1-4y}dy = \frac{t}{1+t^2}dt$ then by integration at both sides respectively, one can get

$\ln (1-4y)(-{1 \over 4}) = \ln(1+ t^2)({1\over2}) + C$ then

$\ln (1-4y) = -2\ln(1+ t^2) +e^C $

$1-4y = e^{-2\ln(1+ t^2)+e^C}$ =

$y = -{1\over4}e^{-2\ln(1+ t^2)+e^C} + 1/4$ then with a condition $y(1) = 1/4$

$-{1\over4}e^{-2\ln(1+ t^2)+e^C} = 0 $ But RHS always $>0$..

Beverlie
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1 Answers1

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Here I think going for the homogeneous/particular solution method is straightforward since the particular solution $y=\frac 14$ is really trivial.

So for homegeneous equation you have $\displaystyle \frac {y'}y=\frac{-4t}{1+t^2}$ which gives $\displaystyle \ln|y|=A-2\ln(1+t^2)$.

So the general solution is $\displaystyle y=\underbrace{\frac B{(1+t^2)^2}}_\text{homogeneous sol}+\underbrace{\frac 14}_\text{particular sol}$

The (quite silly) initial condition requires $y(1)=\frac B4+\frac 14=\frac 14$ so $B=0$ and $y=\frac 14$ is the final solution.

Note that with your method (which is equally good) you arrive at:

$\ln|1-4y|=-2\ln(1+t^2)+C\iff 1-4y=\frac{D}{(1+t^2)^2}$ and the initial condition also gives $D=0$ and finally $y=\frac 14$.

In your work you added $C$ it disappeared and reappeared by magic as an additive constant later. But in fact you should take the exponential of each side and this lead to a multiplicative constant $D=\pm e^C$, with $\pm$ due to the absolute value on LHS.

Also since $y=0$ is always a solution of an homogeneous equation, we generally go for an arbitrary multiplicative constant $D$ in place of $\pm e^C\cup \{0\}$.

zwim
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  • first thx for the great comment. second, I had edited OP following your advice that I cannot just arbitrarily make the Constant appear and disappear. However, it seems obvious that my approach not working with some unknown reason(to me.) any comment for the edited OP? – Beverlie Jun 18 '17 at 06:24
  • It should be $\ln|1-4y|=-2\ln(1+t^2)+C$ then $\displaystyle 1-4y=\pm e^{-2\ln(1+t^2)+C}=\pm e^C\times e^{-2\ln(1+t^2)}$. Now it is true that $e^C\neq 0$ but this is an homogeneous linear ODE, so we should apply the Cauchy theorem which tells us that a general solution is a constant D times $\exp(\int ...)$, which takes into account that $y=0$ is always a possible solution, rather than our little solutions cooking. – zwim Jun 18 '17 at 06:30
  • I see.. so above reasoning works to abandon all other possibilities except y = 0.. for homgenous part lol – Beverlie Jun 18 '17 at 06:39