2 Differential Equations by Substitution

In summary, the person is not familiar with solving differential equations, but has made some progress. They don't know how to solve the first problem, but know how to solve the second. They know that the roots of the characteristic equation are $\pm\sqrt{m}i$ and that the general solution will be the two-parameter family $(z,t)$.
  • #1
abhay1
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0
solve the following differential equation with the suggested change of variables.

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  • #2
What progress have you made so far on these DE's? Do you know how to start solving them?
 
  • #3
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:

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

FAQ: 2 Differential Equations by Substitution

What is the concept of substitution in differential equations?

Substitution in differential equations is a method of solving equations by replacing one variable with another. This allows for the equation to be simplified and solved for the new variable. In the case of 2 differential equations, substitution involves replacing both variables with new ones to create a system of equations that can be solved simultaneously.

How does substitution help in solving differential equations?

Substitution allows for the simplification of complex equations, making them easier to solve. In 2 differential equations, substitution helps in creating a system of equations that can be solved simultaneously, providing a solution for both equations at the same time.

Can substitution be used for all types of differential equations?

Substitution can be used for many types of differential equations, including separable, linear, and exact equations. However, it may not always be the most efficient or effective method for solving certain equations. It is important to consider other methods as well when approaching a differential equation problem.

What are the steps involved in solving 2 differential equations by substitution?

The steps for solving 2 differential equations by substitution are:1. Determine the variables to be substituted and the new variables to be used.2. Use the chain rule to express the derivatives in terms of the new variables.3. Substitute the new variables into the original equations.4. Solve the resulting system of equations.5. Use the solution to find the values of the original variables.

Are there any limitations to using substitution in solving differential equations?

Yes, there are limitations to using substitution in solving differential equations. It may not always be possible to find suitable substitutions for all variables, and in some cases, the resulting equations may still be difficult to solve. Additionally, substitution may not always provide the most accurate or precise solution compared to other methods such as numerical approximation or using computer software.

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