Dy/dx=(3y^2+x^2)/(2xy)? Solve as homogeneous differential equation?
dy/dx = (3y^2+x^2) / 2xy 2xy dy = (3y^2+x^2) dx Let y= vx dy = v dx + x dv 2xy dy = (3y^2+x^2) dx 2x (vx) (x dv + v dx) = (3(vx)^2 + x^2 ) dx 2x^3v dv + 2x^2v^2 dx = 3x^2v^2 dx + x^2 dx 2x^3 v dv =(x^2v^2 +x^2) dx 2x^3 v dv = x^2(v^2+1) dx 2v dv /(v^2+1) = 1/x dx ∫ 2v dv /(v^2+1) = ∫1/x dx -------------------- (1) Let u = v^2+1 du = 2v dv ∫ 2v dv /(v^2+1) = ∫ du/ u = ln(u) = ln(v^2+1) (1) becomes: ln (v^2+1) = ln (x) + C1 v^2+1 = e^ln(x) e^C1 v^2+1 = Cx v^2 = Cx-1 y^2 / x^2 = Cx-1 y^2 = x^2(Cx-1) y = ± x sqrt (Cx-1)
Homogeneous Differential Equation?
(3x^2 - y^2)dx + (xy - x^3y^-1)dy = 0 I have a real bad habit of occasionally clamming up my math skills and won't get anywhere for hours on a single problem. This is currently one of those problems. I understand you need to put it into a ratio v = y/x and work from there. But every time I do the integral is unsolvable. Any help?
Dy/dx=x^2 +3y^2/2xy?Solve as homogeneous differential equation?
i) Taking in proper brackets, your question suppose to be: dy/dx = (x² + 3y²)/2xy ii) Let y = vx and differentiating, dy/dx = v + x*(dv/dx) iii) ==> v + x*(dv/dx) = (1 + 3v²)/2v This simplifies as: (2v)dv/(1 + v²) = dx/x Integrating, ln|1 + v²| = ln|x| + C Substituting v = y/x and simplifying, Ans: (x² - y²) = C*(x³) EDIT: @Rapidfir: Can't you read the solution before marking TD? It is atrocious! Clearly in first line I stated "Taking in proper brackets, your question suppose to be: dy/dx = (x² + 3y²)/2xy" and provided the solution accordingly. It is up to the questioner to decide whether it is correct or wrong. Invariably about 95% of the questioners are unaware of presenting the questions with proper brackets. Only those who are answering may have to reasonably read the question, understand and provide the solution. Further this question is quite common in this topic and definitely it is not presented with proper brackets. Further, for your kind information, I am in teaching line for more than 4 decades and very much aware of 'BODMAS' rule. My approach in teaching is: 'Don't discourage anyone for wrong doing; but try to correct them with proper guidance'.
Could Someone Please Solve differential equation y dx + (2x + 3y)dy = 0?
By flipping the derivative this turns into a first-order linear differential equation. Find the general solution by separating the variables then integrating: y dx + (2x + 3y) dy = 0 y + (2x + 3y)(dy / dx) = 0 (2x + 3y)(dy / dx) = -y dy / dx = -y / (2x + 3y) dx / dy = -(2x + 3y) / y dx / dy = -2x / y - 3 dx / dy + 2x / y = -3 dx / dy + P(y)x = f(y) P(y) = 2 / y f(y) = -3 I(y) = ℮^[∫ P(y) dy] I(y) = ℮^(∫ 2 / y dy) I(y) = ℮^(2ln|y|) I(y) = ℮^(lny²) I(y) = y² I(y)x = ∫ I(y)f(y) dy xy² = ∫ -3y² dy xy² = -y³ + C xy² = C - y³ x = C / y² - y
Solve following homogeneous equation: (2x+3y)(dx) +(y-x)(dy)=0 use y=vx?
(2x + 3y) (dx) + (y - x) (dy) = 0 Let y = vx => dy = vdx + xdv => (2x + 3vx) dx + (vx - x) (vdx + xdv) = 0 => (2 + 3v) dx + (v - 1) (vdx + xdv) = 0 => (2 + 3v + v^2 - v) dx + x (v - 1) dv = 0 => dx / x+ (v - 1) dv / (v^2 + 2v + 2) = 0 => ∫dx / x + (1/2)∫(2v + 2 - 4) dv / [(v + 1)^2 + 1] = 0 => ln l x l + (1/2) ∫d(v^2 + 2v + 2) / (v^2 + 2v + 2) - 2∫dv / [(v + 1)^2 + 1] = 0 => ln l x l + (1/2) ln (v^2 + 2v + 2) - 2tanֿ¹(v + 1) = c/2 => ln x^2 + ln (v^2 + 2v + 2) - 4tan ֿ¹(v + 1) = c => ln (x^2v^2 + 2vx^2 + 2x^2) - 4tanֿ¹(v + 1) = c => ln (y^2 + 2xy + 2x^2) - 4tanֿ¹[(x + y)/x] = c.
Find general solution of differential equation: y'=(3y^2-x^2) / 2xy?
This is a homogeneous differential equation. It can be reduced to variable separable form by making the substitution, y=vx, so that, dy/dx=v+x dv/dx---- (1) Keep the above equations in mind. So, divide the numerator and denominator of your equation's RHS by x^2 => dy/dx=(3(y/x)^2 - 1)/2(y/x) apply (1) on this equation. => v+x(dv/dx)=(3v^2-1)/2v => x(dv/dx)=(v^2-1)/2v separate the variables and integrate both sides => log|v^2-1|=log|x|+log|c| => v^2=1+xc => (y/x)^2=1+xc This is the solution of your differential equation.