MHB 2 Differential Equations by Substitution

[abhay1]

solve the following differential equation with the suggested change of variables.

View attachment 6279

 

[Ackbach]

What progress have you made so far on these DE's? Do you know how to start solving them?

 

[abhay1]

actually i don't know how to solve them.Can anyone help please.

 

[Ackbach]

Sure! So on the first one, the suggested substitution is $x=e^t$. The reason this works is because the original DE is a Cauchy-Euler equation, where you're multiplying each derivative by successively lower powers of $x$. So, if we do this substitution, we need all the derivatives in place (note here that $y'=dy/dx$ and $\dot{y}=dy/dt$):
\begin{align*}
x&=e^t \\
\frac{dx}{dt}&=e^t \\
\frac{dt}{dx}&=e^{-t} \\
y'&=\frac{dy}{dx}=\frac{dy}{dt} \, \frac{dt}{dx} = \dot{y} \, e^{-t} \\
y''&=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} \, \frac{dy}{dx} = \left[\frac{d}{dt} \, \frac{dy}{dx}\right] \frac{dt}{dx}
=\frac{d}{dt}\left[ \dot{y} \, e^{-t} \right] e^{-t}
=\left[ \ddot{y} e^{-t} - \dot{y} e^{-t} \right] e^{-t} = e^{-2t} (\ddot{y}-\dot{y}).
\end{align*}
Now, substitute all this stuff into the original DE. What do you get?

 

[MarkFL]

For the second problem, we are given:

$$\left(1+x^2\right)^2\frac{d^2y}{dx^2}+2x\left(1+x^2\right)\frac{dy}{dx}+my=0$$

We are told to use the substitutions:

$$x=\tan(t)\implies \frac{dx}{dt}=\sec^2(t)$$

$$y(x)=z(t)$$

In the second substitution, if we implicitly differentiate w.r.t $t$, we obtain:

$$\frac{dy}{dx}\cdot\frac{dx}{dt}=\frac{dz}{dt}$$

Using the implication from the first substitution, there results:

$$\frac{dy}{dx}=\cos^2(t)\frac{dz}{dt}$$

Now, if we differentiate again w.r.t $t$, we have:

$$\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}=\cos^2(t)\frac{d^2z}{dt^2}-2\cos(t)\sin(t)\frac{dz}{dt}$$

Again, using the implication from the first substitution, there results:

$$\frac{d^2y}{dx^2}=\cos^4(t)\frac{d^2z}{dt^2}-2\cos^3(t)\sin(t)\frac{dz}{dt}$$

Now, let's plug all of the substitutions and their implications into the given ODE:

$$\left(1+\tan^2(t)\right)^2\left(\cos^4(t)\frac{d^2z}{dt^2}-2\cos^3(t)\sin(t)\frac{dz}{dt}\right)+2\tan(t)\left(1+\tan^2(t)\right)\left(\cos^2(t)\frac{dz}{dt}\right)+mz=0$$

Next, let's use the Pythagorean identity $1+\tan^2(\theta)=\sec^2(\theta)$:

$$\sec^4(t)\left(\cos^4(t)\frac{d^2z}{dt^2}-2\cos^3(t)\sin(t)\frac{dz}{dt}\right)+2\tan(t)\sec^2(t)\left(\cos^2(t)\frac{dz}{dt}\right)+mz=0$$

Distribute:

$$\frac{d^2z}{dt^2}-2\tan(t)\frac{dz}{dt}+2\tan(t)\frac{dz}{dt}+mz=0$$

Combine like terms:

$$\frac{d^2z}{dt^2}+mz=0$$

Can you proceed?

 

[MarkFL]

Just to follow up with the second problem, givent that $m$ is a positive real number, we see the roots of the characteristic, or auxiliary equation are:

$$r=\pm\sqrt{m}i$$

And so, by the theory of linear homogeneous equations, we know the general solution will be the two-parameter family:

$$z(t)=c_1\cos(\sqrt{m}t)+c_2\sin(\sqrt{m}t)$$

Back-substituting for $z$ and $t$, we obtain:

$$y(x)=c_1\cos\left(\sqrt{m}\arctan(x)\right)+c_2\sin\left(\sqrt{m}\arctan(x)\right)$$