1. Method: Reduction to homogeneous type

A non-homogeneous equation can sometimes be transformed into a homogeneous equation by an appropriate substitution.

To solve a differential equation of the form

We can reduce it to a homogeneous form by changing variables. The variables xand y are replaced with X and Y related by the equations:

where h and k are constants chosen to make the given equation homogeneous.

Steps to follow when solving:

1. Substitute x = X + h \ and y = Y + k \:

   Replacingx and y in the differential equation:

   \frac{d(Y + k)}{d(X + h)} = \frac{a(X + h) + b(Y + k) + c}{A(X + h) + B(Y + k) + C}  \

2. Differentiate the Substituted Expression:

   Differentiate Y + k \ with respect to X + h \:

   \frac{d(Y + k)}{d(X + h)} = \frac{dY}{dX}  \

   Since \frac{d}{d(X + h)} \ is the same as \frac{d}{dX} \, we have:

    \frac{dY}{dX}  = \frac{dy}{dx}

3. Substitute and simplify:

   Substitute x and y into the equation:

   \frac{dY}{dX} = \frac{a(X + h) + b(Y + k) + c}{A(X + h) + B(Y + k) + C} \

   Distribute the constants:

   \frac{dY}{dX} = \frac{a(X + h) + b(Y + k) + c}{A(X + h) + B(Y + k) + C} \

4. Adjust Constants to Achieve Homogeneity:

   To make the equation homogeneous, ensure that the terms involving h and k cancel out. This requires:

   a h + b k + c = 0 \

    A h + B k + C = 0  \

5. Resulting Homogeneous Equation:

   Once h and k are chosen to satisfy these conditions, the simplified equation will be:

   \frac{dY}{dX} = \frac{aX + bY}{AX + BY}  \

By substituting x = X + h \ and y = Y + k \ into the original equation and adjusting the constants h andk to satisfy the conditions, you transform the differential equation into a homogeneous form. This helps in solving the equation using methods suited for homogeneous differential equations.

Now, applying the previous methods <in>, we can find the solution:

Substitute Y = vX

\frac{dY}{dX} = v + X \frac{dv}{X}

So,

v + X \frac{dv}{dX} = \frac{aX + bvX}{AX+BvX}

Now we have a separable equation,

X\frac{dv}{dX} = \frac{a + bv}{A +Bv} - v

Then solve the equation in terms of v, X. Then, substitute v = \frac{Y}{X} in the solution. Finally, substitute X = x - h and Y = y - k.

2. Examples

Example 1:

Solve the following differential equation:

Step 1: Substitute x = X + h\ and y = Y + k\:

\frac{d(Y + k)}{d(X + h)} = \frac{2(X + h) - 5(Y + k) + 2}{3(X + h) + 4(Y + k) - 7}

Step 2: Differentiate the substituted expression:

Since \frac{d(Y + k)}{d(X + h)} = \frac{dY}{dX}\, we can rewrite the equation as:

\frac{dY}{dX} = \frac{2(X + h) - 5(Y + k) + 2}{3(X + h) + 4(Y + k) - 7}

Step 3: Substitute and simplify:

Expanding both the numerator and denominator gives:

\frac{dY}{dX} = \frac{2X + 4h - 5Y - 5k + 2}{3X + 3h + 4Y + 4k - 7}

Step 4: Adjust constants to achieve homogeneity:

We want the terms involving h and k to cancel out, so we solve:

2h - 5k + 2 = 0\

3h + 4k - 7 = 0\

Solving this system:

h = \frac{27}{23}, \quad k = \frac{20}{23}\

Step 5: Substitute h and k and simplify:

Substitute h = \frac{27}{23} and k = \frac{20}{23}:

\frac{dY}{dX} = \frac{4X - 5Y}{3X + 4Y}\

Now, the equation is in homogeneous form.

Substitute: Y = vX

Then,

\frac{dY}{dX} = v + X \frac{dv}{dX}

v + X \frac{dv}{dX} = \frac{2X - 5vX}{3X + 4vX}

X\frac{dv}{dX} = \frac{2+5v}{3+4v} -v

X\frac{dv}{dX}= \frac{2 + 2v - 4v^2}{3 + 4v}
\frac{3 + 4v}{2(1 + v - 2v^2)} dv = \frac{1}{X} dX

Now integrating both sides:

\int \frac{3 + 4v}{2(1 + v - 2v^2)} dv = \int \frac{1}{X} dX

Partial fractions on the left-hand side:

\int \frac{1}{3(2v+1)} - \frac{7}{6(v-1)} dv = \int \frac{1}{X} dX

\frac{1}{6} \ln{|2v+1|} - \frac{7}{6} \ln{|v-1|} = \ln{|X|} + c

Resubstituting v = \frac{Y}{X}

\frac{1}{6} \ln{|2\frac{Y}{X}+1|} - \frac{7}{6} \ln{|\frac{Y}{X}-1|} = \ln{|X|} + c

Now substituting: X = x - h = x - \frac{27}{23} and Y = y - k = y - \frac{20}{23}

Finally,

\frac{1}{6} \ln{\left|2\frac{y - \frac{20}{23}}{x - \frac{27}{23}}+1\right|} - \frac{7}{6} \ln{\left|\frac{y - \frac{20}{23}}{x - \frac{27}{23}}-1\right|} = \ln{\left|x - \frac{27}{23}\right|} + c

Example 2:

Given the equation, transform it into homogeneous form:

Step 1: Substitute x = X + h\ and y = Y + k\:

\frac{d(Y + k)}{d(X + h)} = \frac{3(X + h) + 2(Y + k) - 5}{(X + h) - (Y + k) + 1}\

Step 2: Differentiate the substituted expression:

Since \frac{d(Y + k)}{d(X + h)} = \frac{dY}{dX}\, we can rewrite the equation as:

\frac{dY}{dX} = \frac{3(X + h) + 2(Y + k) - 5}{(X + h) - (Y + k) + 1}\

Step 3: Substitute and simplify:

Expanding both the numerator and denominator:

Numerator:

3(X + h) + 2(Y + k) - 5 = 3X + 3h + 2Y + 2k - 5\

Denominator:

(X + h) - (Y + k) + 1 = X + h - Y - k + 1\

This gives:

\frac{dY}{dX} = \frac{3X + 3h + 2Y + 2k - 5}{X + h - Y - k + 1}\

Step 4: Adjust constants to achieve homogeneity:

We want the terms involving h and kto cancel out, so we solve:

3h + 2k - 5 = 0\

h - k + 1 = 0\

Solving this system:

h = \frac{3}{5}, \quad k = \frac{8}{5}

Step 5: Substitute h and k and simplify:

Substitute h = \frac{3}{5} and k = \frac{8}{5}:

\frac{dY}{dX} = \frac{3X + 2Y}{X - Y}\

Now, the equation is in homogeneous form.

Example 3:

Example 4:

Example 5:

3. Practice Questions

Solve the following non-homogeneous equation:

1.    \frac{dy}{dx} = \frac{y - x + 1}{y + x - 5}\

2.    \frac{dy}{dx} = \frac{x - 2y + 3}{2x - 4y + 5}\

Not sure about your answers? Don't worry, click the link below to check the video where the answers are explained.

Practice Questions Discussion