Answer to Question #242756 in Calculus for JaytheCreator

Question

Differentiate the following functions (i) If 𝑦 = 𝑒 −3𝑡 sin 4𝑡, prove that 𝑑 2𝑦 𝑑𝑥 2 + 6 𝑑𝑦 𝑑𝑥 + 2𝑦 = 0. (ii) Given that sin(𝑥 2 + 𝑦) = 𝑦 2 (3𝑥 + 1), show that 𝑑𝑦 𝑑𝑥 = 2𝑥 cos(𝑥 2 + 𝑦) − 3𝑦 2 2𝑦(3𝑥 + 1) − cos(𝑥 2 + 𝑦

Answer 1

(i) First, we know that $𝑦 = 𝑒^{−3𝑡} \sin{(4𝑡)}$, then we proceed to derive

$\\ \frac{dy}{dt}=\frac{d}{dt}(𝑒^{−3𝑡}\sin{(4𝑡)}) \\ \frac{dy}{dt}=\sin{(4𝑡)}\frac{d(𝑒^{−3𝑡} )}{dx}+𝑒 ^{−3𝑡} \frac{d(\sin{(4𝑡)})}{dx} \\ \frac{dy}{dt}=\sin{(4𝑡)}(-3𝑒^{−3𝑡})+𝑒^{−3𝑡}(4\cos{(4t)}) \\ \frac{dy}{dt}=-3𝑒^{−3𝑡}\sin{(4𝑡)}+4𝑒^{−3𝑡}\cos{(4t)} \\ \frac{dy}{dt}=𝑒^{−3𝑡} \Big[ 4\cos{(4t)} -3\sin{(4𝑡)} \Big]$

After the first derivative we can work on the second one:

$\frac{d^2y}{dt^2}=\frac{d}{dt}\Big(𝑒^{−3𝑡} [ 4\cos{(4t)} -3\sin{(4𝑡)} ] \Big) \\ \frac{d^2y}{dt^2}=𝑒^{−3𝑡} \frac{d}{dt}\Big( 4\cos{(4t)} -3\sin{(4𝑡)} \Big)+ ( 4\cos{(4t)} -3\sin{(4𝑡)} ) \frac{d}{dt}\Big( 𝑒^{−3𝑡}\Big) \\ \frac{d^2y}{dt^2}=𝑒^{−3𝑡} \big( -16\sin{(4t)} -12\cos{(4𝑡)} \big)+ ( 4\cos{(4t)} -3\sin{(4𝑡)} )\big(-3𝑒^{−3𝑡}\big) \\ \frac{d^2y}{dt^2}=𝑒^{−3𝑡} \big( -16\sin{(4t)} -12\cos{(4𝑡)} -12\cos{(4t)} +9\sin{(4𝑡)} \big) \\ \frac{d^2y}{dt^2}=𝑒^{−3𝑡} \big( -7\sin{(4t)} -24\cos{(4𝑡)}\big)$

Now we have

$𝑦 = 𝑒^{−3𝑡} \sin{(4𝑡)} \\ \frac{dy}{dt}=𝑒^{−3𝑡} \Big[ 4\cos{(4t)} -3\sin{(4𝑡)} \Big] \\ \frac{d^2y}{dt^2}=-𝑒^{−3𝑡} \big( 7\sin{(4t)} +24\cos{(4𝑡)}\big)$

And we can substitute this on the DE, $\frac{d^2y}{dt^2}+6\frac{dy}{dt}+2y=0$, to find if this is true for y:

$\\ 𝑒^{−3𝑡} \big( -7\sin{(4t)} -24\cos{(4𝑡)}\big)+6\Big(𝑒^{−3𝑡} \big[ 4\cos{(4t)} -3\sin{(4𝑡)} \big]\Big)+2 𝑒^{−3𝑡} \sin{(4𝑡)}=0 \\ 𝑒^{−3𝑡} \Bigg( -7\sin{(4t)} -24\cos{(4𝑡)}+ 24\cos{(4t)} -18\sin{(4𝑡)} +2\sin{(4𝑡)} \Bigg)=0$

$\\ \implies -𝑒^{−3𝑡} \Bigg( 23\sin{(4𝑡)} \Bigg) \neq 0$

That implies that $𝑦 = 𝑒^{−3𝑡} \sin{(4𝑡)}$ is not a solution for the DE $\frac{d^2y}{dt^2}+6\frac{dy}{dt}+2y=0$.

(ii) Now we have two expressions, A and B:

$\sin(x^2 + 𝑦) = y^2(3x+ 1) \\ \implies A=B$

We proceed to differentiate both terms and we have to use implicit differentiation:

$A'=\cos(x^2 + y)(2x +\frac{dy}{dx}) \\ A'=2x\cos(x^2 + y) +\frac{dy}{dx}(\cos(x^2 + y))$

$B'= 3y^2+2y(3x+ 1)\frac{dy}{dx} \\ B'= 3y^2+2y(3x+ 1)\frac{dy}{dx}$

Now we substitute to find $\frac{dy}{dx}$ :

$A'=B' \\2x\cos(x^2 + y) -3y^2= (2y(3x+ 1)-\cos(x^2 + y))\frac{dy}{dx} \\ \text{ } \\ \implies \cfrac{dy}{dx} = \cfrac{2x\cos(x^2 + y) -3y^2}{ 2y(3x+ 1)-\cos(x^2 + y)}$

That last result is consistent with the expected expression.

Reference:

• Thomas, G. B., & Finney, R. L. (1961). Calculus. Addison-Wesley Publishing Company.