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Solve (x+3)²y"-4(x+3)y'+6y = log(x+3)

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best_answer

We are given the differential equation:
\[
(x+3)^2 y'' - 4(x+3) y' + 6y = \log(x+3)
\]

{Step 1: Change of variable.}

Let \( t = x + 3 \). Then \( \frac{d}{dx} = \frac{d}{dt} \), so the equation becomes:
\[
t^2 \frac{d^2 y}{dt^2} - 4t \frac{dy}{dt} + 6y = \log t
\]

This is a linear non-homogeneous second-order differential equation with variable coefficients.

{Step 2: Solve the homogeneous equation.}

\[
t^2 y'' - 4t y' + 6y = 0
\]

This is a Cauchy-Euler equation. Assume a solution of the form \( y = t^r \). Then:
\[
y' = r t^{r-1}, \quad y'' = r(r-1)t^{r-2}
\]

Substitute into the equation:
\[
t^2 [r(r-1)t^{r-2}] - 4t [r t^{r-1}] + 6t^r = 0
\]

\[
r(r-1)t^r - 4r t^r + 6t^r = 0
\Rightarrow t^r [r(r-1) - 4r + 6] = 0
\]

\[
r^2 - 5r + 6 = 0 \Rightarrow (r-2)(r-3) = 0
\Rightarrow r = 2, 3
\]

So the general solution of the homogeneous equation is:
\[
y_h(t) = C_1 t^2 + C_2 t^3
\]

{Step 3: Find a particular solution.}

We try variation of parameters. The homogeneous solutions are:
\[
y_1 = t^2, \quad y_2 = t^3
\]

Compute the Wronskian:
\[
W = 
\begin{vmatrix}
t^2 & t^3 \\
2t & 3t^2
\end{vmatrix}
= t^2 \cdot 3t^2 - t^3 \cdot 2t = 3t^4 - 2t^4 = t^4
\]

Using variation of parameters:

\[
y_p = u_1(t) y_1 + u_2(t) y_2
\]

Where:
\[
u_1' = -\frac{y_2 \cdot f(t)}{W} = -\frac{t^3 \cdot \log t}{t^4} = -\frac{\log t}{t}
\Rightarrow u_1 = -\int \frac{\log t}{t} dt = -\frac{(\log t)^2}{2}
\]

\[
u_2' = \frac{y_1 \cdot f(t)}{W} = \frac{t^2 \cdot \log t}{t^4} = \frac{\log t}{t^2}
\Rightarrow u_2 = \int \frac{\log t}{t^2} dt
\]

Integrate by parts:
\[
\int \frac{\log t}{t^2} dt,\quad u = \log t,\ dv = t^{-2} dt,\ 
du = \frac{1}{t} dt,\ v = -t^{-1}
\]

\[
= -\frac{\log t}{t} + \int \frac{1}{t^2} dt = -\frac{\log t}{t} - \frac{1}{t}
\]

So:
\[
u_2 = -\frac{\log t}{t} - \frac{1}{t}
\]

Therefore, the particular solution is:
\[
y_p = u_1 y_1 + u_2 y_2 = -\frac{(\log t)^2}{2} t^2 + \left(-\frac{\log t}{t} - \frac{1}{t} \right)t^3
\]

\[
= -\frac{1}{2} t^2 (\log t)^2 - t^2 \log t - t^2
\]

{Final solution:}
\[
y(t) = y_h + y_p = C_1 t^2 + C_2 t^3 - \frac{1}{2} t^2 (\log t)^2 - t^2 \log t - t^2
\]

{Now substitute back } \( t = x + 3 \):

\[
\boxed{
y(x) = C_1 (x+3)^2 + C_2 (x+3)^3 - \frac{1}{2} (x+3)^2 (\log(x+3))^2 - (x+3)^2 \log(x+3) - (x+3)^2
}
\]
 

Posted by

Saumya Singh

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