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Original 2026-01-01
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1 <p>Euler’s theorem states that, if f(x, y, z) is a homogeneous function of degree n, then: x∂f∂x + y∂f∂y + z∂f∂z = n f(x, y, z) This means that if we multiply each variable by its partial derivative and then add them, we will get the degree n times the original function. </p>
1 <p>Euler’s theorem states that, if f(x, y, z) is a homogeneous function of degree n, then: x∂f∂x + y∂f∂y + z∂f∂z = n f(x, y, z) This means that if we multiply each variable by its partial derivative and then add them, we will get the degree n times the original function. </p>
2 <p>When a function is homogeneous, scaling its variable by a<a>number</a>t changes the whole function by tn. Euler’s theorem is a rule that comes from this scaling behavior. The steps given below explain Euler’s theorem:</p>
2 <p>When a function is homogeneous, scaling its variable by a<a>number</a>t changes the whole function by tn. Euler’s theorem is a rule that comes from this scaling behavior. The steps given below explain Euler’s theorem:</p>
3 <p>Step 1: Assume f(x, y, z) is homogeneous of degree n. f(tx, ty, tz) = tn f(x, y, z)</p>
3 <p>Step 1: Assume f(x, y, z) is homogeneous of degree n. f(tx, ty, tz) = tn f(x, y, z)</p>
4 <p>Step 2: Differentiate both sides of the<a>equation</a>with respect to t, applying the derivative rule to each term.</p>
4 <p>Step 2: Differentiate both sides of the<a>equation</a>with respect to t, applying the derivative rule to each term.</p>
5 <p>Step 3: Apply the chain rule to the left-hand side: x∂f∂x + y∂f∂y + z∂f∂z</p>
5 <p>Step 3: Apply the chain rule to the left-hand side: x∂f∂x + y∂f∂y + z∂f∂z</p>
6 <p>Step 4: Differentiate the right-hand side, ntn - 1f(x, y, z)</p>
6 <p>Step 4: Differentiate the right-hand side, ntn - 1f(x, y, z)</p>
7 <p>Step 5: Set t = 1, and you get, x∂f∂x + y∂f∂y + z∂f∂z = n f(x, y, z)</p>
7 <p>Step 5: Set t = 1, and you get, x∂f∂x + y∂f∂y + z∂f∂z = n f(x, y, z)</p>
8 <p>Hence, Euler's theorem is proved.</p>
8 <p>Hence, Euler's theorem is proved.</p>
9 <p><strong>Homogeneous Differential Equation from Homogeneous Function</strong></p>
9 <p><strong>Homogeneous Differential Equation from Homogeneous Function</strong></p>
10 <p>A homogeneous differential equation whose right-hand side is made from a homogeneous function of x and y. It can be written as:</p>
10 <p>A homogeneous differential equation whose right-hand side is made from a homogeneous function of x and y. It can be written as:</p>
11 <p>dydx = f(x, y) If f(x, y) is a homogeneous function, then the equation is referred to as a homogeneous differential equation.</p>
11 <p>dydx = f(x, y) If f(x, y) is a homogeneous function, then the equation is referred to as a homogeneous differential equation.</p>
12 <p>A function f(x, y) is called homogeneous of degree n if we can write it as:</p>
12 <p>A function f(x, y) is called homogeneous of degree n if we can write it as:</p>
13 <p>f(x, y) = xn × gyx or f(x, y) = yn × gxy</p>
13 <p>f(x, y) = xn × gyx or f(x, y) = yn × gxy</p>
14 <p>The homogeneous function depends on the<a>ratio</a>of y to x or x to y.</p>
14 <p>The homogeneous function depends on the<a>ratio</a>of y to x or x to y.</p>
15 <p>A differential equation is called homogeneous when it can be written in the form of, dydx = g(x, y) This means the right side of the equation depends only on the ratio yx. The steps to solve it are provided below.</p>
15 <p>A differential equation is called homogeneous when it can be written in the form of, dydx = g(x, y) This means the right side of the equation depends only on the ratio yx. The steps to solve it are provided below.</p>
16 <p><strong>Step 1:</strong>We use a new variable: v = yx or y = v . x</p>
16 <p><strong>Step 1:</strong>We use a new variable: v = yx or y = v . x</p>
17 <p><strong>Step 2:</strong>If y = v . x, we use the<a>product</a>rule to find dydx</p>
17 <p><strong>Step 2:</strong>If y = v . x, we use the<a>product</a>rule to find dydx</p>
18 <p>dydx = v + xdvdx</p>
18 <p>dydx = v + xdvdx</p>
19 <p><strong>Step 3:</strong>Substitute the value dydx into the given equation:</p>
19 <p><strong>Step 3:</strong>Substitute the value dydx into the given equation:</p>
20 <p>v + xdvdx = g(v)</p>
20 <p>v + xdvdx = g(v)</p>
21 <p><strong>Step 4:</strong>Rearrange to isolate dvdx</p>
21 <p><strong>Step 4:</strong>Rearrange to isolate dvdx</p>
22 <p>xdvdx = g(v) - v</p>
22 <p>xdvdx = g(v) - v</p>
23 <p><strong>Step 5:</strong>Separate the variables to rewrite the equation to separate v and x:</p>
23 <p><strong>Step 5:</strong>Separate the variables to rewrite the equation to separate v and x:</p>
24 <p>dvg(v) - v = dxx</p>
24 <p>dvg(v) - v = dxx</p>
25 <p><strong>Step 6: I</strong>ntegrate both sides,</p>
25 <p><strong>Step 6: I</strong>ntegrate both sides,</p>
26 <p>1g(v) - vdv =1xdx + C</p>
26 <p>1g(v) - vdv =1xdx + C</p>
27 <p> Here, C is the constant.</p>
27 <p> Here, C is the constant.</p>
28 <p><strong>Step 7:</strong>Since v = yx, we substitute it back into the solution to get the answer in terms of x and y. If the equation is dxdy = f(x, y) is homogeneous, we use x = vy and solve it the same way. </p>
28 <p><strong>Step 7:</strong>Since v = yx, we substitute it back into the solution to get the answer in terms of x and y. If the equation is dxdy = f(x, y) is homogeneous, we use x = vy and solve it the same way. </p>
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