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Homogeneous Differential Equation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Homogeneous Differential Equation is considered one the most difficult concept.

  • 30 Questions around this concept.

Solve by difficulty

Lety = y(x) be the solution of the differential equation \left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1 Then 6 y^2(e) is equal to

The solution of the differential equation \frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0 is

Let \mathrm{y=y(x)} be the solution of the differential equation \mathrm{\left(3 y^2-5 x^2\right) y \mathrm{~d} x+2 x\left(x^2-y^2\right) \mathrm{d} y=0}such that \mathrm{y(1)=1}. Then \mathrm{\left|(y(2))^3-12 y(2)\right|} is equal to 

The slope of tangent at any point (x, y) on a curve y = y(x) is   \frac{x^{2}+y^{2}}{2xy}\, ,\, x>0  If y (2) = 0, then a value of y (8) is:

The solution curve of the differential equation $y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$ passing through the point $(\mathrm{e}, 1)$ is

Concepts Covered - 2

Homogeneous Differential Equation

A function f(x, y) is said to be a homogeneous function of degree n if it satisfies the property

\\\mathrm{\mathit{f(\lambda x,\lambda y)=\lambda^nf(x,y)}}

Consider the following examples

\\\mathrm{1.\;\;\;\;f(x,y)=x^3-4xy^2}\\\mathrm{2.\;\;\;\;f(x,y)=x-3y}\\\mathrm{3.\;\;\;\;f(x,y)=\tan\frac{x}{y}}\\\\\mathrm{in\;the \;above\;examples\;if\;we\;replace\;x\;and\;y\;with}\\\mathrm{\lambda x\;and\;\lambda y,\;where\;\lambda\;is\;non-zero,\;we\;get }\\\\\mathrm{1.\;\;\;\;f(\lambda x,\lambda y)=\left (\lambda x \right )^3-4(\lambda x)(\lambda y)^2=\lambda^3(x^3-4xy^2)=\lambda^3f(x,y)}\\\mathrm{2.\;\;\;\;f(\lambda x,\lambda y)=\lambda x-3(\lambda y)=\lambda(x-3y)=\lambda f(x,y)}\\\mathrm{3.\;\;\;\;f(\lambda x,\lambda y)=\tan \frac{\lambda x}{\lambda y}=\tan \frac{x}{y}=\lambda ^0f(x,y)}

Now, if the function is given as 
\\\mathrm{4.\;\;\;\;f(x,y)=\sin x+\cos y, then\,\,f(\lambda x, \lambda y)\neq\lambda^nf(x,y) }\\\\\text{Observe that it is possible to write examples 1,2 and 3 in the }\\\mathrm{form\;of\;f(\lambda x,\lambda y)=\lambda^nf(x,y).}\\\mathrm{But\;example\;4\;can't\;be\;written\;in\;this\;form.}\\\text{Here,\;example\;1,\;2\;and 3 are homogeneous equation of degree 3, 1\;and 0}\\\text{respectively and example 4 is not a homogeneous function.}

We can define homogeneous differential equation as follows : 

\\\mathrm{Any \;DE\; of\; the\; form\; \mathit{M(x,y)dx+N(x,y)dy=0} \;or\;\mathit{\frac{d y}{d x}=-\frac{M(x, y)}{N(x, y)}}\;}\\\mathrm{is\;called\;homogeneous\;if\;\mathit{M(x,y)}\;and\;\mathit{N(x,y)}\;are\;homogeneous}\\\text {functions of the same degree. }

\\\mathrm{Since,\;M(x,y)\;and\;N(x,y)\;are\;both\;homogeneous\;function\;of}\\\text{degree }n,\;\text{ then DE can be reduced to a function of y/x}\\ \\\frac{d y}{d x}=-\frac{M(x, y)}{N(x, y)}=\phi\left(\frac{y}{x}\right)

This equation can be solved by the substitution y = vx.

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y=vx}\\\mathrm{\Rightarrow\;\;\;\;\;\;\;\;\;\;\; \frac{dy}{dx}=v+x\frac{dv}{dx}}\\\text { Thus, } \frac{d y}{d x}=\phi\left(\frac{y}{x}\right) \text { transforms to }\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; v+x\frac{dv}{dx}=\phi\left ( v \right )}\\\\\Rightarrow\;\;\;\;\;\;\;\;\;\;\; \frac{d v}{\phi(v)-v}=\frac{d x}{x}\\\text{The variables have now been separated and the solution is}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;}\int \frac{\mathrm{d} \mathrm{v}}{\phi(\mathrm{v})-\mathrm{v}}=\ln \mathrm{x}+\mathrm{c}\\\\\text{After the integration v should be replaced by y/x} \\\text{to get the required solution.}

Reducible to Homogeneous Form

If the differential equation is of the form
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ax}+\mathrm{by}+\mathrm{c}}{\mathrm{d} \mathrm{x}+\mathrm{e} \mathrm{y}+\mathrm{f}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1)

It can be reduced to a homogeneous differential equation as follows:

Put x = X + h, y = Y + k                         ……. (2)

where X and Y are new variables and h and k are constants yet to be chosen

From (2)

dx = dX, dy = dY

Equation (1), thus reduces to

\mathrm{\frac{dY}{dX}=\frac{a(X+h)+b(Y+k)+c}{d(X+h)+e(Y+k)+f}=\frac{aX+bY+(ah+bk+c)}{dX+eY+(dh+ek+f)}\;\;\;\;\;\;\;\;\ldots(3)}

In order to have equation (3) as a homogeneous differential equation, choose h and k such that the following equations are satisfied :

\left.\begin{array}{rl}{a h+b k+c} & {=0} \\ {d h+e k+f} & {=0}\end{array}\right\}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(4)

Now, (3) becomes

\frac{d Y}{d X}=\frac{a X+b Y}{d X+e Y}

which is a homogeneous differential equation and can be solved by putting Y = vX.

 

Note:

If d/a = e/b (=t say) , the above method does not apply.

In such cases, Equation (1) becomes

\frac{d y}{d x}=\frac{a x+b y+c}{t(a x+b y)+f}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(6)

which can be solved by putting ax + by = v

Separate the variables and integrate to get the required solution.

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Homogeneous Differential Equation
Reducible to Homogeneous Form

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Reducible to Homogeneous Form

Mathematics for Joint Entrance Examination JEE (Advanced) : Algebra

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