Chapter 4: Q35E (page 200)

Given that $1 + t, 1 + 2 t$ , and $1 + 3 t^{2}$ are solutions to the differential equation $y'' + p (t) y' + q (t) y = g (t)$ , find the solution to this equation that satisfies $y (1) = 2$ , $y' (1) = 0$ .

Short Answer

Expert verified

The solution to the given equation $y'' + p (t) y' + q (t) y = g (t)$ that satisfies $y (1) = 2$ , $y' (1) = 0$ is $y = 1 + 2 t - t^{2}$ .

Step by step solution

Substitute the value for t

Given differential equation is $y'' + p (t) y' + q (t) y = g (t)$ and solutions are $1 + t, 1 + 2 t$ and $1 + 3 t^{2}$ .

So, the solution is $y = c_{1} (1 + t) + c_{2} (1 + 2 t) + c_{3} (1 + 3 t^{2})$

Now impose the initial conditions are:

$y (1) = c_{1} (1 + 1) + c_{2} (1 + 2) + c_{3} (1 + 3) 2 c_{1} + 3 c_{2} + 4 c_{3} = 2 . . . (1)$

And

$y' (t) = c_{1} + 2 c_{2} + 6 c_{3} t y' (1) = c_{1} + 2 c_{2} + 6 c_{3} c_{1} + 2 c_{2} + 6 c_{3} = 0$

Finding c1,c2,c3

By the trial-and-error method, we will get $c_{1} = \frac{2}{3}, c_{2} = \frac{2}{3}, c_{3} = - \frac{2}{3}$

So, the solution is $\begin{array}{rcl} \frac{2}{3} (1 + t) + \frac{2}{3} (1 + 2 t) - \frac{1}{3} (1 + 3 t^{2}) & = & \frac{2}{3} + \frac{2}{3} - \frac{1}{3} + \frac{2 t}{3} + \frac{4 t}{3} - t^{2} \\ = & 1 + 2 t - t^{2} \end{array}$

Therefore, the solution is $y = 1 + 2 t - t^{2}$ .

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Given that $1 + t, 1 + 2 t$ , and $1 + 3 t^{2}$ are solutions to the differential equation $y'' + p (t) y' + q (t) y = g (t)$ , find the solution to this equation that satisfies $y (1) = 2$ , $y' (1) = 0$ .

Short Answer

Step by step solution

Substitute the value for t

Finding c1,c2,c3

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Given that 1+t,1+2t, and1+3t2are solutions to the differential equationy''+p(t)y'+q(t)y=g(t), find the solution to this equation that satisfiesy1=2,y'(1)=0.

Short Answer

Step by step solution

Substitute the value for t

Finding c1,c2,c3

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Calculus

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Given that $1 + t, 1 + 2 t$ , and $1 + 3 t^{2}$ are solutions to the differential equation $y'' + p (t) y' + q (t) y = g (t)$ , find the solution to this equation that satisfies $y (1) = 2$ , $y' (1) = 0$ .