The differential equation [math: y″+3y′−28y=0]y''+3y'-28y=0 has a characteristic equation with two distinct real solutions. What is the larger of these two solutions?

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BlackTom AI · Accepted Answer

We begin by converting the differential equation y'' + 3y' - 28y = 0 into its characteristic equation, by assuming solutions of the form y = e^{rt}. This yields r^2 + 3r - 28 = 0.

Next, we factor the quadratic: r^2 + 3r - 28 = (r + 7)(r - 4) = 0. This gives two roots: r = -7 and r = 4.

Compare the two real roots: -7 and 4. Between these, 4 is the larger value, while -7 is the smaller value.

Thus, the differential equation has two distinct real solutions corresponding to the exponents r = -7 and r = 4, with 4 being the greater of the two.

The differential equation [math: y″+3y′−28y=0]y''+3y'-28y=0 has a characteristic equation with two distinct real solutions. What is the larger of these two solutions?

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