# Thread: 2 Differential Equations by Substitution

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

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

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

4. 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?

5. For the second problem, we are given:

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

We are told to use the substitutions:

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

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

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

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

Using the implication from the first substitution, there results:

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

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

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

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

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

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

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

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

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

Distribute:

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

Combine like terms:

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

Can you proceed?

6. 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:

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

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

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

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

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