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Differential Equations - Solution of homogeneous differential equations of first order and first degree

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A function f(x,y)f(x, y) is said to be a homogeneous function of degree nn if f(λx,λy)=λnf(x,y)f(\lambda x, \lambda y) = \lambda^n f(x, y) for any non-zero constant λ\lambda. Visually, if you imagine a point (x,y)(x, y) moving along a line passing through the origin, the function scales uniformly according to the distance from the origin raised to the power nn.

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A differential equation of the form dydx=F(x,y)\frac{dy}{dx} = F(x, y) is called homogeneous if F(x,y)F(x, y) is a homogeneous function of degree zero. This means F(x,y)F(x, y) can be expressed entirely as a function of the ratio yx\frac{y}{x} or xy\frac{x}{y}. On a graph, this implies that the slope of the solution curve is constant along any straight line y=mxy = mx passing through the origin.

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The primary method to solve a homogeneous differential equation of the form dydx=g(yx)\frac{dy}{dx} = g(\frac{y}{x}) is the substitution y=vxy = vx. By differentiating this with respect to xx using the product rule, we get dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}. This substitution transforms the equation into a variable separable form.

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After substituting y=vxy = vx and dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}, the differential equation becomes v+xdvdx=g(v)v + x\frac{dv}{dx} = g(v), which can be rearranged to dvg(v)−v=dxx\frac{dv}{g(v) - v} = \frac{dx}{x}. This represents a transformation from a complex slope field to a simpler one where variables vv and xx can be integrated independently.

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If the differential equation is given in the form dxdy=h(xy)\frac{dx}{dy} = h(\frac{x}{y}), it is more efficient to use the substitution x=vyx = vy. In this case, dxdy=v+ydvdy\frac{dx}{dy} = v + y\frac{dv}{dy}. This is particularly useful when the expression for dxdy\frac{dx}{dy} is simpler than that of dydx\frac{dy}{dx}.

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Once the integration in terms of vv and xx (or vv and yy) is complete, the final step is to substitute v=yxv = \frac{y}{x} (or v=xyv = \frac{x}{y}) back into the general solution to obtain the relation between xx and yy. The resulting integral often involves logarithmic functions or inverse trigonometric functions, representing families of curves such as spirals or dilated geometric shapes.

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To verify if an equation is homogeneous, replace xx with λx\lambda x and yy with λy\lambda y. If λ\lambda cancels out completely (leaving λ0\lambda^0), the equation is homogeneous of degree zero. Geometrically, this confirms that the direction of the tangent to the solution curve depends only on the angle of the position vector, not its magnitude.

📐Formulae

Homogeneous Function: f(λx,λy)=λnf(x,y)\text{Homogeneous Function: } f(\lambda x, \lambda y) = \lambda^n f(x, y)

Standard Form: dydx=F(x,y)=g(yx)\text{Standard Form: } \frac{dy}{dx} = F(x, y) = g\left(\frac{y}{x}\right)

Substitution 1: y=vx  ⟹  dydx=v+xdvdx\text{Substitution 1: } y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}

Separated Form 1: dvg(v)−v=dxx\text{Separated Form 1: } \frac{dv}{g(v) - v} = \frac{dx}{x}

Substitution 2: x=vy  ⟹  dxdy=v+ydvdy\text{Substitution 2: } x = vy \implies \frac{dx}{dy} = v + y\frac{dv}{dy}

General Solution: ∫dvg(v)−v=∫dxx+C\text{General Solution: } \int \frac{dv}{g(v) - v} = \int \frac{dx}{x} + C

💡Examples

Problem 1:

Solve the differential equation: x2dydx=x2−2y2+xyx^2 \frac{dy}{dx} = x^2 - 2y^2 + xy

Solution:

Step 1: Express dydx\frac{dy}{dx} in terms of xx and yy: dydx=x2−2y2+xyx2=1−2(yx)2+yx\frac{dy}{dx} = \frac{x^2 - 2y^2 + xy}{x^2} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x} Step 2: Substitute y=vxy = vx and dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}: v+xdvdx=1−2v2+vv + x\frac{dv}{dx} = 1 - 2v^2 + v Step 3: Simplify and separate variables: xdvdx=1−2v2  ⟹  dv1−2v2=dxxx\frac{dv}{dx} = 1 - 2v^2 \implies \frac{dv}{1 - 2v^2} = \frac{dx}{x} Step 4: Integrate both sides: ∫dv1−(2v)2=∫dxx\int \frac{dv}{1 - (\sqrt{2}v)^2} = \int \frac{dx}{x} 122ln⁡∣1+2v1−2v∣=ln⁡∣x∣+C\frac{1}{2\sqrt{2}} \ln \left| \frac{1 + \sqrt{2}v}{1 - \sqrt{2}v} \right| = \ln |x| + C Step 5: Substitute v=yxv = \frac{y}{x}: 122ln⁡∣x+2yx−2y∣=ln⁡∣x∣+C\frac{1}{2\sqrt{2}} \ln \left| \frac{x + \sqrt{2}y}{x - \sqrt{2}y} \right| = \ln |x| + C

Explanation:

We first confirm the equation is homogeneous by dividing through by x2x^2. We then use the standard substitution y=vxy=vx to convert it into a variable separable form and integrate using the partial fraction formula for ∫dxa2−x2\int \frac{dx}{a^2-x^2}.

Problem 2:

Solve: xdydx=y−xtan⁡(yx)x \frac{dy}{dx} = y - x \tan\left(\frac{y}{x}\right)

Solution:

Step 1: Write in the form dydx=g(yx)\frac{dy}{dx} = g(\frac{y}{x}): dydx=yx−tan⁡(yx)\frac{dy}{dx} = \frac{y}{x} - \tan\left(\frac{y}{x}\right) Step 2: Substitute y=vxy = vx and dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}: v+xdvdx=v−tan⁡(v)v + x\frac{dv}{dx} = v - \tan(v) Step 3: Simplify: xdvdx=−tan⁡(v)  ⟹  dvtan⁡(v)=−dxxx\frac{dv}{dx} = -\tan(v) \implies \frac{dv}{\tan(v)} = -\frac{dx}{x} cot⁡(v)dv=−dxx\cot(v) dv = -\frac{dx}{x} Step 4: Integrate: ∫cot⁡(v)dv=−∫dxx\int \cot(v) dv = -\int \frac{dx}{x} ln⁡∣sin⁡v∣=−ln⁡∣x∣+ln⁡C\ln |\sin v| = -\ln |x| + \ln C ln⁡∣sin⁡v∣=ln⁡∣Cx∣  ⟹  sin⁡v=Cx\ln |\sin v| = \ln \left| \frac{C}{x} \right| \implies \sin v = \frac{C}{x} Step 5: Substitute v=y/xv = y/x: sin⁡(yx)=Cx or xsin⁡(yx)=C\sin\left(\frac{y}{x}\right) = \frac{C}{x} \text{ or } x \sin\left(\frac{y}{x}\right) = C

Explanation:

This problem uses the trigonometric ratio within the homogeneous function. The substitution y=vxy=vx allows the vv terms to cancel out conveniently, leading to a simple integration of the cotangent function.