Answer to Question #90800 in Differential Equations for Paul Anthony

Question

Solve the following pDEs
A). d²u÷dx²= 8xy²+1
B). d²u÷dxy- du÷dx=6xe^x

Answer 1

A)

$$\frac{\partial^2u}{\partial x^2}= 8xy^2+1.$$

Integrate each side of the equation by x. Antiderivatives will differ by an arbitrary function C₁(y):

$$\frac{\partial u}{\partial x}= 4x^2y^2+x+C_1(y).$$

Repeat this:

$$u = \frac{4}{3}x^3y^2+\frac{x^2}{2}+C_1(y)x +C_2(y).$$

C₂(y) is an arbitrary function.

B)

$$\frac{\partial ^2u}{\partial x \partial y} -\frac{\partial u}{\partial x}=6xe^x.$$

Integrate each side of the equation by x. Antiderivatives will differ by an arbitrary function C₁(y):

$$\frac{\partial u}{\partial y} - u = 6(x-1)e^x+C_1(y).$$

This is linear ordinary differential linear equation by y. Then by the well-known formula

[https://en.wikipedia.org/wiki/Linear_differential_equation#First-order_equation_with_variable_coefficients]

$$u = e^y\left(\int\limits_{y_0}^{y}{\Bigl(6(x-1)e^x + C_1(t)\Bigr)e^{-t}dt} + C_2(x) \right),$$

where y₀ is any preselected number (let y₀=0 for example), C₂(x) is an arbitrary function of x.

Then

$$u = \left(6(x-1)e^x\int\limits_{y_0}^{y}{e^{-t}dt} + \int\limits_{y_0}^{y}{C_1(t)e^{-t}dt} + C_2(x)\right) e^y.$$

Now let the integrals not depend on x: this could be taken into account in arbitrary C₂(x). Then consider a new arbitrary function C₃(y) based on arbitrary function C₁(y):

$$C_3(y) = e^y\int\limits_{y_0}^{y}{C_1(t)e^{-t}dt},$$

and a new arbitrary function C₄(x) based on arbitrary function C₂(x):

$$\int\limits_{0}^{y}{e^{-t}dt} = -e^{-y} + 1,$$ $$C_4(x) = 6(x-1)e^x + C_2(x).$$

Then the expression is simplified:

$$u = C_4(x)e^y + C_3(y) - 6(x-1)e^x.$$