Answer to Question #118565 in Differential Equations for Anderson Joseph

Question

Give the complete solutions of the twoo problems

1.7) Find a+b+c if phi (x) = a(e^x) + b(e^cx) is a solution to the initial value problem: second derivative of y + first derivative of y - 2y = 0; y(0) = 2, first derivative of y(0) = 1. Here a,b, and c are constants.

1.8) Find the positive constant a(>0) so that the parametric equation: x = at, y= -t^2 is a solution to y= x(dy/dx) + (dy/dx)^2.

Answer 1

1.7) Given $\phi(x) = a e^x + b e^{cx}$ is solution of $y''+y' - 2y = 0, y(0)=2,y'(0)=1$.

Hence, $y = ae^x+be^{cx} \implies y' = ae^x+bce^{cx} \ and \ y'' = ae^x+bc^2e^{cx}$.

So, $(ae^x+bc^2e^{cx}) + (ae^x+bce^{cx}) - 2(ae^x+be^{cx}) = 0$

$\implies b(c^2+c-2) = 0$ _____________(1)

Also, $y(0)=2, y'(0)=1$

$\implies a+b=2$ __________________(2)

and $a+bc = 1$ ___________________(3)

From equation(1) if $b = 0$ , then from equation(2) $a = 2$ and from equation (3) $a = 1$ which is not possible, so $b\neq 0$.

Now, so from equation(1), $c^2+c-2 = 0 \implies (c+2)(c-1) = 0$

$\implies either \ c = 1, \ or \ c = -2.$

If $c = 1,$ then from equation (3) $a+b = 1$ which contradict the equation (2), so $c \neq 1$.

Only $c = -2$ is possible, so from equation (3) $a-2b = 1$ _____________(4).

So, by equation (2) - equation(3), we get $c = \frac{1}{3}$ and hence $a = \frac{5}{3}$ .

So, $a+b+c=2+(-2) = 0$.

1.8) Given $x = at, y= -t^2$ is solution of $y= x(dy/dx) + (dy/dx)^2$.

Now, $\frac{dx}{dt} = a, \frac{dy}{dt} = -2t \implies \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{2t}{a}$.

By putting in given differential equation, we get

$-t^2 = at (\frac{-2t}{a}) + (\frac{-2t}{a})^2 \\ \implies -a^2 t ^2= -2 a^2 t^2 + 4 t^2 \implies a^2 = 4 \\ \implies a = 2$