# Finding the Particular Solution for a Second-Order Differential Equation

• Mastur
In summary, the homework equation is x^2y\frac{d^2y}{dx^2} + (x\frac{dy}{dx} - y)^2 = 0. The problem requires for the 2nd derivative, which makes it really hard (In my part). If you're only solving for the constants, you do that using the two equations y'(1)=1 and y(1)=2. You only need the expressions for y(x) and y'(x) to write down those two equations.
Mastur

## Homework Statement

$x^2y\frac{d^2y}{dx^2} + (x\frac{dy}{dx} - y)^2 = 0$

$y^2=C_{1}x^2+C_{2}x$

$y=2;y'=1 when x=1$

## Homework Equations

Can someone give me the answer? Just the answer, the particular solution. Please? I'm having really hard time calculating for it. This is one of the seatworks given to us, and this is the remaining problem that I cannot answer.

## The Attempt at a Solution

differentiating the second equation
$y'=\frac{2C_{1}x+C_{2}}{2y}$
$y''=\frac{4C_{1}y-2C_{1}^2x^2+C_{2}^2+3C_{1}C_{2}x}{4y^2}$
I'm stuck right there. I substituted the y' and y'' to the first equation but the answer seems not equal to zero. Is my differentiation right?

Last edited:
The numerator of the first derivative isn't correct. Perhaps a typo? The second derivative looks wrong too.

Is the point of the problem to just find the constants c1 and c2? If so, you only need to calculate the first derivative.

I see, I forgot to place the x right after the C1? Is that what you're talking about? Fixed for you.

Why first derivative only? The problem (first equation) requires for the 2nd derivative, which makes it really hard (In my part).

If you're only solving for the constants, you do that using the two equations y'(1)=1 and y(1)=2. You only need the expressions for y(x) and y'(x) to write down those two equations.

If you're also supposed to verify that y(x) is indeed a solution to the differential equation, you will need y''(x).

I see. I substituted the values y'=1; y=2 when x=1, and this is what I got.

$C_{1}=0$

$C_{2}=4$

Then, the particular solution should be:
$y=2\sqrt{x}$

Am I correct?

Yes, that function satisfies the differential equation and the initial conditions.

Never thought that would be as easy as that. @_@

Thank you!

## What is a differential equation?

A differential equation is a mathematical equation that relates the rate of change of a function to the function itself. It is used to model and solve problems that involve rates of change, such as growth and decay, motion, and electrical circuits.

## What is the difference between an ordinary and a partial differential equation?

An ordinary differential equation (ODE) involves only one independent variable, while a partial differential equation (PDE) involves multiple independent variables. ODEs are used to model systems with one variable, such as population growth, while PDEs are used to model systems with multiple variables, such as heat transfer.

## How are differential equations solved?

Differential equations can be solved analytically or numerically. Analytical solutions involve finding an exact formula for the function, while numerical solutions involve using mathematical algorithms and computers to approximate the solution.

## What are initial conditions and boundary conditions in a differential equation?

Initial conditions are values of the dependent variable at a specific starting point, while boundary conditions are values of the dependent variable at specific points along the domain of the independent variable. These conditions are used to determine the unique solution to a differential equation.

## What are some real-life applications of differential equations?

Differential equations are used in many fields, including physics, engineering, economics, and biology. They are used to model and understand natural phenomena, such as population growth, chemical reactions, and fluid flow. They also play a crucial role in the development of technologies, such as control systems and image processing.

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