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7.4 Workshop Exercises 4

For more worked examples, see Gilbert and Jordan Guide to Mathematical Methods, pages 153–164.

The main themes this week are: double integration (W4.1-4.2); 2nd order homogeneous differential equations, constant coefficients (W4.3-4.4), integrable differential equations (W4.5-4.6); separable differential equations (W4.7-4.8); linear first-order differential equations (W4.9-4.10). For questions W4.3-4.4, see slides 4.45-49 in the MATH101 notes.

W4.1. Calculate the repeated integrals

J=∫12∫01(y⁢x3+y2⁢ex)⁢d⁢x⁢d⁢yJ=\int_{1}^{2}\int_{0}^{1}(yx^{3}+y^{2}e^{x})\,dxdy

and

K=∫01∫12(y⁢x3+y2⁢ex)⁢d⁢y⁢d⁢x.K=\int_{0}^{1}\int_{1}^{2}(yx^{3}+y^{2}e^{x})\,dydx.

In each case, start by sketching the region of integration.

W4.2. Evaluate the double integral ∫0π6∫0π6sec2⁡(x+y)⁢d⁢x⁢d⁢y\int_{0}^{\frac{\pi}{6}}\int_{0}^{\frac{\pi}{6}}\sec^{2}(x+y)\,dx\,dy.

W4.3. Find the general solutions to the following differential equations:

a) d2⁢yd⁢x2+4⁢d⁢yd⁢x+3⁢y=0\frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+3y=0.

b) d2⁢yd⁢x2+4⁢d⁢yd⁢x+4⁢y=0\frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+4y=0.

c) d2⁢yd⁢x2+4⁢d⁢yd⁢x+5⁢y=0\frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+5y=0.

W4.4. Solve the following initial value problems:

a) d2⁢yd⁢x2-3⁢d⁢yd⁢x-4⁢y=0\frac{d^{2}y}{dx^{2}}-3\frac{dy}{dx}-4y=0, y⁢(0)=1y(0)=1, y′⁢(0)=9y^{\prime}(0)=9.

b) d2⁢yd⁢x2+2⁢d⁢yd⁢x+10⁢y=0\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+10y=0, y⁢(0)=0y(0)=0, y′⁢(0)=1y^{\prime}(0)=1.

W4.5.Find the general solution of the differential equation (x+1)⁢d⁢yd⁢x=x+2(x+1)\frac{dy}{dx}=x+2.

W4.6. Solve the initial-value problem

(x3+x-2)2⁢d⁢yd⁢x=3⁢x2+1, y⁢(0)=32.(x^{3}+x-2)^{2}\frac{dy}{dx}=3x^{2}+1,\;\;y(0)=\frac{3}{2}.

(*) What is the maximum possible range of values of xx on which this defines a valid solution?

W4.7. Find the general solution of the separable differential equation d⁢yd⁢x=y⁢(x2+1)\frac{dy}{dx}=y(x^{2}+1). Find the particular solution with y⁢(0)=3y(0)=3.

W4.8. Find the general solution to the differential equation d⁢yd⁢x=x3⁢(y+1)2\frac{dy}{dx}=\frac{x}{3(y+1)^{2}}.

W4.9. Solve the initial value problem (x3+1)⁢d⁢yd⁢x-3⁢x2⁢y=x2(x^{3}+1)\frac{dy}{dx}-3x^{2}y=x^{2}, y⁢(0)=1y(0)=1.

W4.10.Put the differential equation

2⁢d⁢yd⁢x=2⁢x+2⁢y⁢tan⁡x2\frac{dy}{dx}=2x+2y\tan x

into the standard form of a first-order linear equation, and find the general solution.