Question

3. Find the general solution of the partial differential equation

6uxx −5uxy −6uyy =52e^3x+2y

Answer 1

1 STEP: We rewrite this equation using the notation

$$\frac{\partial}{\partial x}\equiv D\quad\text{and}\quad\frac{\partial}{\partial y}=D'\longrightarrow\\[0.3cm] 6u_{xx}-5u_{xy}-6u_{yy}=52\cdot e^{3x+2y}\longrightarrow\\[0.3cm] \boxed{\left(6D^2-5DD'-6D'^2\right)u=52\cdot e^{3x+2y}}$$

2 STEP: Factor the resulting square expression

Note: I will use the letters $x$ and $y$ so that there is no confusion between the operators $D$ , $D'$ and the designation of the discriminant $D$ :

$$6D^2-5DD'-6D'^2\longrightarrow 6x^2-5yx-6y^2=0\\[0.3cm] \left.\begin{array}{l} a=6\\ b=-5y\\ c=-6y^2 \end{array}\right| D=b^2-4ac=(-5y)^2-4\cdot 6\cdot(-6y)\\[0.3cm] D=25y^2+144y^2=169y^2\longrightarrow\sqrt{D}=13y\\[0.3cm] \left[\begin{array}{l} x_1=\displaystyle\frac{-b-\sqrt{D}}{2a}=\displaystyle\frac{5y-13y}{12}=-\displaystyle\frac{2y}{3}\\[0.3cm] x_2=\displaystyle\frac{-b+\sqrt{D}}{2a}=\displaystyle\frac{5y+13y}{12}=\displaystyle\frac{3y}{2} \end{array}\right.\\[0.3cm] 6x^2-5xy-6y=6\cdot\left(x-\left(-\frac{2y}{3}\right)\right)\cdot\left(x-\frac{3y}{2}\right)\\[0.3cm] \boxed{6D^2-5DD'-6D'^2=6\cdot\left(D+\frac{2}{3}\cdot D'\right)\cdot\left(D-\frac{3}{2}\cdot D'\right)}$$

3 STEP: A bit of theory (without proof)

Suppose we need to solve a partial differential equation

$$f\left(D,D'\right)u=\varphi(x,y)$$

The solution consists of two parts:

$$u(x,y)=C.F.+P.I.,\,\,\text{where}\\[0.31cm] \left[\begin{array}{l} C.F.-\text{the complementary function}\\[0.3cm] P.I.-\text{Particular Integral } \end{array}\right.$$

To find the complementary function (C.F.)

Put $D=m$ and $D'=1$ in $f\left(D, D'\right)=0$ , then we get an equation, which is called the auxillary equation.

∴ the auxillary equation is $f(m,1)=0$ i.e.,

$$a_0m^n + a_1m^{n−1} + · · · + a_n = 0$$

Let the roots of this equation be $m_1$ , $m_2$ , . . . , $m_n$ . If the roots are real (or imaginary) and distinct, then

$$C.F.=f_1(y+m_1x)+f_2(y+m_2x)+ · · · +f_n(y+m_nx)$$

To find Particular Integral (P.I.)

If

$$\varphi(x, y)=e^{ax+by}$$

then

$$P.I.=\frac{1}{f\left(D,D'\right)}\cdot e^{ax+by}=\frac{1}{f(a,b)}\cdot e^{ax+by},\text{provided} f(a, b)\neq0.$$

If $f(a,b)=0$ , then

$$P.I.=x\cdot\frac{1}{f'\left(D, D'\right)}\cdot e^{ax+by},\\[0.3cm] \text{where}\,\,\,f'\left(D, D'\right)\,\,\text{is the partial derivative of} \,\,f\left(D, D'\right) w.r.t \,\,D.$$

4 STEP: Put theory into practice

$$С.F. :\left(D+\frac{2}{3}\cdot D'\right)\cdot\left(D-\frac{3}{2}\cdot D'\right)u=0\longrightarrow\\[0.3cm] \boxed{C.F.=f_1\left(y-\frac{2x}{3}\right)+f_2\left(y+\frac{3x}{2}\right)}\\[0.3cm] P.I. : P.I.=\frac{1}{\left(D+\frac{2}{3}\cdot D'\right)\cdot\left(D-\frac{3}{2}\cdot D'\right)}\cdot\frac{52}{6}\cdot e^{3x+2y}\\[0.3cm] P.I.=\frac{1}{\left(3+\frac{2}{3}\cdot 2\right)\cdot\left(3-\frac{3}{2}\cdot 2\right)}\cdot\frac{52}{6}\cdot e^{3x+2y}\\[0.3cm] P.I.=\frac{1}{\left(3+\frac{4}{3}\right)\cdot\underbrace{\left(3-3\right)}}_{=0}\cdot\frac{52}{6}\cdot e^{3x+2y}\\[0.3cm]$$

This means that you must first differentiate the expression $\left(6D^2-5DD'-6D'^2\right)$ by $D$ , so

$$\frac{\partial}{\partial D}\left(6D^2-5DD'-6D'^2\right)=12D-5D'$$

Then,

$$P.I. : P.I.=\frac{1}{12D-5D'}\cdot 52\cdot e^{3x+2y}\\[0.3cm] P.I.=\frac{1}{12\cdot3-5\cdot 2}\cdot 52\cdot e^{3x+2y}\\[0.3cm] P.I.=\frac{52\cdot e^{3x+2y}}{26}\equiv 2\cdot e^{3x+2y}\\[0.3cm]$$

Conclusion,

$$u(x,y)=C.F.+P.I.\longrightarrow\\[0.3cm] \boxed{u(x,y)=f_1\left(y-\frac{2x}{3}\right)+f_2\left(y+\frac{3x}{2}\right)+2\cdot e^{3x+2y}}$$

ANSWER

$$u(x,y)=f_1\left(y-\frac{2x}{3}\right)+f_2\left(y+\frac{3x}{2}\right)+2\cdot e^{3x+2y}$$