Question

solve the given initial value problem. Give the largest interval I over which the solution is defined.

(1+t^2)dy/dx + x = tan^-1 t , x(0) = 4

[Hint: In your solution let u = tan^-1 t ]

Answer 1

Answer on Question #70256 – Math – Differential Equations

Question

Solve the given initial value problem. Give the largest interval I over which the solution is defined.

$$(1 + t^2) \frac{dx}{dt} + x = \tan^{-1} t, \quad x(0) = 4$$

[Hint: In your solution let $u = \tan^{-1} t$ ]

Solution

$$(1 + t^2) \frac{dx}{dt} + x = \tan^{-1} t$$

This linear first order ODE can be re-arranged to give the following standard form

$$\frac{dx}{dt} + \frac{1}{1 + t^2} x = \frac{\tan^{-1} t}{1 + t^2}$$

where $P(t) = \frac{1}{1 + t^2}$ and $Q(t) = \frac{\tan^{-1} t}{1 + t^2}$

The ODE can be solved using the integrating factor method.

$$IF = e^{\int P(t) dt} = e^{\int \frac{1}{1 + t^2} dt} = e^{\tan^{-1} t}$$

This factor is defined so that the equation becomes equivalent to:

$$\frac{d}{dt} (IF \cdot x) = IF \cdot Q(x)$$

Integrating both sides with respect to $t$, gives:

$$x = \frac{1}{IF} \int IF \cdot Q(x) \, dt$$

Then

$$x = \frac{1}{e^{\tan^{-1} t}} \int e^{\tan^{-1} t} \cdot \frac{\tan^{-1} t}{1 + t^2} \, dt$$

Let $u = \tan^{-1} t$. Then

$$du = \frac{1}{1 + t^2} \, dt$$ $$\begin{aligned} \int e^{\tan^{-1} t} \cdot \frac{\tan^{-1} t}{1 + t^2} \, dt &= \int e^u u \, du = u e^u - \int e^u \, du = u e^u - e^u + C \\ &= e^{\tan^{-1} t} \tan^{-1} t - e^{\tan^{-1} t} + C \\ \int w \, dv &= w v - \int v \, dw \\ w &= u, \quad dw = du \\ dv &= e^u du, \quad v = e^u \\ x &= \frac{1}{e^{\tan^{-1} t}} \left( e^{\tan^{-1} t} \tan^{-1} t - e^{\tan^{-1} t} + C \right) \end{aligned}$$ $$x = \tan^{-1} t - 1 + \frac{C}{e^{\tan^{-1} t}}$$ $$x(0) = 4 \Rightarrow \tan^{-1}(0) - 1 + \frac{C}{e^{\tan^{-1}(0)}} = 4$$ $$0 - 1 + \frac{C}{e^{0}} = 4$$ $$C = 5$$

Then

$$x = \tan^{-1} t - 1 + \frac{5}{e^{\tan^{-1} t}}$$

The solution is defined on $(-\infty, \infty)$.

Answer: $x = \tan^{-1} t - 1 + \frac{5}{e^{\tan^{-1} t}}, t \in (-\infty, \infty)$.

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