Related Formula
(dy)/(dx) = f(ax+by+c)$$\frac{dy}{dx} = f(ax+by+c)$$
Substitute t = ax+by+c$t = ax+by+c$ to reduce the equation to variable separable form.
Core Logic
The differential equation can be written as:
(dy)/(dx) = -(2x+3y-2)/(4x+6y-7)$$\frac{dy}{dx} = -\frac{2x+3y-2}{4x+6y-7}$$
Observe that 4x+6y = 2(2x+3y)$4x+6y = 2(2x+3y)$. Let us substitute t = 2x+3y-2$t = 2x+3y-2$.
Taking derivatives with respect to x$x$:
(dt)/(dx) = 2 + 3(dy)/(dx) ⇒ (dy)/(dx) = (1)/(3)((dt)/(dx) - 2)$$\frac{dt}{dx} = 2 + 3\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{3}\left(\frac{dt}{dx} - 2\right)$$
Step 1: Translating and Simplifying
Substitute t$t$ into the differential equation:
(1)/(3)((dt)/(dx) - 2) = -(t)/(2(t+2)-7)$$\frac{1}{3}\left(\frac{dt}{dx} - 2\right) = -\frac{t}{2(t+2)-7}$$
(dt)/(dx) - 2 = -(3t)/(2t-3)$$\frac{dt}{dx} - 2 = -\frac{3t}{2t-3}$$
(dt)/(dx) = 2 - (3t)/(2t-3)$$\frac{dt}{dx} = 2 - \frac{3t}{2t-3}$$
(dt)/(dx) = (4t - 6 - 3t)/(2t - 3) = (t - 6)/(2t - 3)$$\frac{dt}{dx} = \frac{4t - 6 - 3t}{2t - 3} = \frac{t - 6}{2t - 3}$$
Step 2: Variable Separation Integration
Separate the variables t$t$ and x$x$:
∫ (2t - 3)/(t - 6) dt = ∫ dx$$\int \frac{2t - 3}{t - 6} dt = \int dx$$
Decompose the fraction algebraically:
∫ (2(t-6) + 9)/(t-6) dt = ∫ ( 2 + (9)/(t-6) ) dt$$\int \frac{2(t-6) + 9}{t-6} dt = \int \left( 2 + \frac{9}{t-6} \right) dt$$
2t + 9ln|t-6| = x + C$$2t + 9\ln|t-6| = x + C$$
Step 3: Restoring Original Variables
Substitute t = 2x + 3y - 2$t = 2x + 3y - 2$ back:
2(2x + 3y - 2) + 9ln|2x + 3y - 2 - 6| = x + C$$2(2x + 3y - 2) + 9\ln|2x + 3y - 2 - 6| = x + C$$
4x + 6y - 4 + 9ln|2x + 3y - 8| = x + C$$4x + 6y - 4 + 9\ln|2x + 3y - 8| = x + C$$
3x + 6y + 9ln|2x + 3y - 8| = C + 4$$3x + 6y + 9\ln|2x + 3y - 8| = C + 4$$
Divide the entire equation by 3:
x + 2y + 3ln|2x + 3y - 8| = C'$$x + 2y + 3\ln|2x + 3y - 8| = C'$$
Step 4: Finding the Constant of Integration
Given initial condition y(0) = 3$y(0) = 3$ (when x=0, y=3$x=0, y=3$):
0 + 2(3) + 3ln|2(0) + 3(3) - 8| = C'$$0 + 2(3) + 3\ln|2(0) + 3(3) - 8| = C'$$
6 + 3ln|1| = C' ⇒ C' = 6$$6 + 3\ln|1| = C' \Rightarrow C' = 6$$
Thus, the specific solution is:
x + 2y + 3ln|2x + 3y - 8| = 6$$x + 2y + 3\ln|2x + 3y - 8| = 6$$
Step 5: Comparing and Final Evaluation
Comparing with the given form α x + β y + 3ln|2x + 3y - γ| = 6$\alpha x + \beta y + 3\ln|2x + 3y - \gamma| = 6$:
α = 1$\alpha = 1$, β = 2$\beta = 2$, γ = 8$\gamma = 8$.
Compute the required expression:
α + 2β + 3γ = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$$\alpha + 2\beta + 3\gamma = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$$
Pattern Recognition
When the coefficients of x and y in the numerator and denominator are proportional (i.e. a₁/a₂ = b₁/b₂$a_1/a_2 = b_1/b_2$), the standard procedure is to use a direct composite substitution t = ax+by$t = ax+by$ which effortlessly maps to a basic logarithmic integral.
Chapter Mix
Class 12 Maths: Differential Equations
