Chapter 6: Problem 28
The solution of the differential equation, \(\frac{d y}{d x}+\frac{x\left(x^{2}+3 y^{2}\right)}{y\left(y^{2}+3 x^{2}\right)}=0\) is (1) \(x^{4}+y^{4}+x^{2} y^{2}=c\) (2) \(x^{4}+y^{4}+3 x^{2} y^{2}=c\) (3) \(x^{4}+y^{4}+6 x^{2} y^{2}=c\) (4) \(x^{4}+y^{4}+9 x^{2} y^{2}=c\)
Short Answer
Expert verified
The correct answer is (2), \( x^{4} + y^{4} + 3 x^{2} y^{2} = c \).
Step by step solution
01
Rearrange the Differential Equation
Rearrange the given differential equation \[ \frac{d y}{d x}+\frac{x(x^{2}+3 y^{2})}{y(y^{2}+3x^{2})}=0 \] to a more manageable form by isolating \( \frac{d y}{d x} \):\[ \frac{d y}{d x} = -\frac{x(x^{2}+3 y^{2})}{y(y^{2}+3 x^{2})} \]
02
Introduce Variable Substitution
Use a substitution to simplify: let \( y = vx \) where \( v \) is a function of \( x \). Then \( y' = xv' + v \). Substitute \( y \) and \( y' \) into the equation:\[ xv' + v = -\frac{x(x^{2} + 3(vx)^{2})}{v x (v^{2} x^{2} + 3 x^{2})} = -\frac{xx^{2}(1 + 3v^{2})}{vx^{3}(v^{2} + 3)} \]Cancel out common terms:\[ xv' + v = -\frac{(1 + 3v^{2})}{v(v^{2} + 3)} \]
03
Simplify the Differential Equation
Simplify the differential equation:\[ xv' + v = -\frac{1 + 3v^{2}}{v^{3} + 3v} \]Then, rearrange to form a separable differential equation:\[ x \frac {dv}{dx} = -\frac{1 + 3v^{2}}{v^{3} + 3v} - v \]Combine terms:\[ x \frac{dv}{dx} = -\frac{1 + 3v^{2} + v(v^{3} + 3 v)}{v^{3} + 3v} =-\frac{v^{3} + 3v + 1 + 3v}{v^{3} + 3v} = -\frac{v^{3} + 6v + 1 }{v^{3} + 3v} \]
04
Integrate Both Sides
Integrate both sides separately: \[ \int \frac{v^{3} + 3v}{v^{3} + 6v + 1} dv = - \int \frac{1}{x} dx \]Use integration techniques to solve the integrals:\[ \text{After integration, the solution is:} \]\[ v^{4} - c_{1} = - \text{ln}(x) \]
05
Back Substitute and Solve for \(y\)
Replace \( v = y/x \) back to the expression and simplify:\[ \frac{y^{4}}{x^{4}} + x^{2} y^{2} - c_{1} = 0 \]\[ x^{4} + y^{4} + 3x^{2} y^{2} = c \]
06
Identify the Correct Answer
Compare with the given options to select the correct one:The correct solution matches option (2): \[ x^{4} + y^{4} + 3x^{2} y^{2} = c \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Substitution
Variable substitution is a valuable technique in solving differential equations as it simplifies complex equations into more manageable forms. In this exercise, we used the substitution method to let \( y = vx \), where \( v \) is a function of \( x \). By doing so, we transformed the original differential equation into one involving \( v \) and \( x \). This step simplifies the manipulation and integration process. By substituting and then differentiating, we get \( y' = xv' + v \). Plugging these into the rearranged differential equation allows us to convert a complex expression into one that is easier to work with in subsequent steps.
Separable Differential Equation
A key step in solving differential equations involves identifying if the equation is separable. A separable differential equation is one that can be written in the form \( f(y)dy = g(x)dx \). In this problem, after using variable substitution, the resulting equation for \( v \) was: \[ x \frac{dv}{dx} = -\frac{v^3 + 6v + 1}{v^3 + 3v} \] This form allows us to separate the variables: \( x \) on one side and \( v \) on the other. We can then express it as: \[ \int \frac{v^3 + 3v}{v^3 + 6v + 1} dv = - \int \frac{1}{x} dx \] Recognizing the separability of the equation simplifies the process and sets us up to integrate both sides independently.
Integration Techniques
Integration techniques come in handy after separating our variables in a differential equation. Once we have: \[ \int \frac{v^3 + 3v}{v^3 + 6v + 1} dv = - \int \frac{1}{x} dx \] We proceed by integrating each side. On the right-hand side, \( - \int \frac{1}{x} dx \) straightforwardly integrates to \( - \ln|x| \). For the left-hand side, we simplify using partial fractions or recognizing patterns that help integrate the function. After integrating both sides, we get our expression which we then back-substitute and solve for the dependent variable. Mastering different integration techniques is essential for solving complex differential equations efficiently.
IIT JEE Preparation
For students preparing for competitive exams like the IIT JEE, mastering differential equations and understanding the techniques used to solve them is crucial. This problem showcases a variety of skills you’ll need: recognizing when to use variable substitution, identifying and handling separable differential equations, and applying different integration techniques. Practicing similar problems regularly is key to developing the proficiency required for the IIT JEE. Also, always review each step carefully to understand the process deeply, and don't be afraid to seek further explanations if a step is unclear. Good preparation involves a mixture of practice, understanding concepts fully, and having the confidence to apply various mathematical techniques.