91Ó°ÊÓ

To find the first derivative of the given function \(y=a\sin(bx+c)\) with respect to \(x\), we can apply the chain rule of differentiation, which states that if \(u(x)\) is a differentiable function, then the derivative of the composition of \(u(x)\) and a scalar multiple, \(ku(x)\), is given by \((ku(x))'=k(u(x))'\). Using this rule, we get: \(y' = \frac{dy}{dx} = a(b\cos(bx+c))\)

Now, let's find the second-order derivative of the given function, which is the derivative of the first derivative with respect to \(x\). We need to apply the chain rule again. Differentiating \(y'\) with respect to \(x\), we get: \(y'' = \frac{d^2y}{dx^2} = ab^2(-\sin(bx+c))\)

The goal is to eliminate the constants \(a\), \(b\), and \(c\) from the second-order derivative equation to arrive at a differential equation that describes the family of curves. Looking at the first derivative, we have: \(y' = a(b\cos(bx+c))\) Dividing both sides by \(a\), we get: \(\frac{y'}{a} = b\cos(bx+c)\) Since we want to eliminate \(a\), \(b\), and \(c\), we can square the equation to get: \(\left(\frac{y'}{a}\right)^2 = b^2\cos^2(bx+c)\) Now, let's square the second derivative equation to eliminate \(a\) and \(b\): \((y'')^2 = a^2b^4\sin^2(bx+c)\) We want to eliminate \(\sin^2(bx+c)\) and \(\cos^2(bx+c)\) to find an expression without \(a\), \(b\), and \(c\). Using the trigonometric identity \(\sin^2(x) + \cos^2(x) = 1\), we get: \(\sin^2(bx+c) = 1 - \cos^2(bx+c)\) Substitute \((1 - \cos^2(bx+c))\) in the second derivative equation: \((y'')^2 = a^2b^4(1 - \cos^2(bx+c))\) Now, we can eliminate the arbitrary constants \(a\), \(b\), and \(c\) by using the relationship we found above, that is: \(\left(\frac{y'}{a}\right)^2 = b^2\cos^2(bx+c)\) From this relation, we can write: \(\frac{(y'')^2}{a^2b^4} = 1 - \frac{(y')^2}{a^2b^2}\) Multiplying both sides with \((a^2b^4)\), we get: \((y'')^2 = a^2b^4 - a^2b^2(y')^2\) Since we only want a differential equation in terms of \(y\), \(y'\), and \(y''\), the final differential equation of the family of curves is: \((y'')^2 + (y')^2 = a^2b^4\) This is the required differential equation that represents the family of curves \(y = a\sin(bx+c)\).