2.1.25 Problem 5

Internal problem ID [18441]
Book : Elementary Differential Equations. By R.L.E. Schwarzenberger. Chapman and Hall. London. First Edition (1969)
Section : Chapter 3. Solutions of first-order equations. Exercises at page 47
Problem number : 5
Date solved : Monday, March 31, 2025 at 05:29:23 PM
CAS classification : [_linear]

Solved using first_order_ode_linear

Time used: 0.040 (sec)

Solve

$$\begin{array}{rl}t{x}^{\prime }+xg\left(t\right)& =h\left(t\right)\end{array}$$

In canonical form a linear first order is

$$\begin{array}{rl}{x}^{\prime }+q(t)x& =p(t)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(t)& =\frac{g\left(t\right)}{t}\\ p(t)& =\frac{h\left(t\right)}{t}\end{array}$$

The integrating factor $\mu$ is

$$\mu ={\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}$$

Therefore the solution is

$$x=(\int \frac{h\left(t\right){\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}}{t}dt+c_1){\mathrm{e}}^{-\int \frac{g\left(t\right)}{t}dt}$$

Summary of solutions found

$$\begin{array}{rl}x& =(\int \frac{h\left(t\right){\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}}{t}dt+c_1){\mathrm{e}}^{-\int \frac{g\left(t\right)}{t}dt}\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 35

ode:=diff(x(t),t)*t+x(t)*g(t) = h(t); 
dsolve(ode,x(t), singsol=all);
 

$$x=(\int \frac{h\left(t\right){\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}}{t}dt+c_1){\mathrm{e}}^{-\int \frac{g\left(t\right)}{t}dt}$$

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & t\left(\frac{d}{dt}x\left(t\right)\right)+x\left(t\right)g\left(t\right)=h\left(t\right)\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dt}x\left(t\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dt}x\left(t\right)=\frac{-x\left(t\right)g\left(t\right)+h\left(t\right)}{t}\\ \bullet & & \text{Collect w.r.t.}x\left(t\right)\text{and simplify}\\ & & \frac{d}{dt}x\left(t\right)=-\frac{g\left(t\right)x\left(t\right)}{t}+\frac{h\left(t\right)}{t}\\ \bullet & & \text{Group terms with}x\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE}\\ & & \frac{d}{dt}x\left(t\right)+\frac{g\left(t\right)x\left(t\right)}{t}=\frac{h\left(t\right)}{t}\\ \bullet & & \text{The ODE is linear; multiply by an integrating factor}\mu \left(t\right)\\ & & \mu \left(t\right)(\frac{d}{dt}x\left(t\right)+\frac{g\left(t\right)x\left(t\right)}{t})=\frac{\mu \left(t\right)h\left(t\right)}{t}\\ \bullet & & \text{Assume the lhs of the ODE is the total derivative}\frac{d}{dt}\left(x\left(t\right)\mu \left(t\right)\right)\\ & & \mu \left(t\right)(\frac{d}{dt}x\left(t\right)+\frac{g\left(t\right)x\left(t\right)}{t})=\left(\frac{d}{dt}x\left(t\right)\right)\mu \left(t\right)+x\left(t\right)\left(\frac{d}{dt}\mu \left(t\right)\right)\\ \bullet & & \text{Isolate}\frac{d}{dt}\mu \left(t\right)\\ & & \frac{d}{dt}\mu \left(t\right)=\frac{\mu \left(t\right)g\left(t\right)}{t}\\ \bullet & & \text{Solve to find the integrating factor}\\ & & \mu \left(t\right)={\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}\\ \bullet & & \text{Integrate both sides with respect to}t\\ & & \int \left(\frac{d}{dt}\left(x\left(t\right)\mu \left(t\right)\right)\right)dt=\int \frac{\mu \left(t\right)h\left(t\right)}{t}dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate the integral on the lhs}\\ & & x\left(t\right)\mu \left(t\right)=\int \frac{\mu \left(t\right)h\left(t\right)}{t}dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}x\left(t\right)\\ & & x\left(t\right)=\frac{\int \frac{\mu \left(t\right)h\left(t\right)}{t}dt+\mathit{C}\mathit{1}}{\mu \left(t\right)}\\ \bullet & & \text{Substitute}\mu \left(t\right)={\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}\\ & & x\left(t\right)=\frac{\int \frac{{\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}h\left(t\right)}{t}dt+\mathit{C}\mathit{1}}{{\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}}\\ \bullet & & \text{Simplify}\\ & & x\left(t\right)={\mathrm{e}}^{-\int \frac{g\left(t\right)}{t}dt}(\int \frac{{\mathrm{e}}^{\int \frac{g\left(t\right)}{t}dt}h\left(t\right)}{t}dt+\mathit{C}\mathit{1})\end{array}$$

✓ Mathematica. Time used: 0.075 (sec). Leaf size: 63

ode=t*D[x[t],t]+x[t]*g[t]==h[t]; 
ic={}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

$$x(t)\to \exp ({\int }_1^t-\frac{g(K[1])}{K[1]}dK[1])({\int }_1^t\frac{\exp (-{\int }_1^{K[2]}-\frac{g(K[1])}{K[1]}dK[1])h(K[2])}{K[2]}dK[2]+c_1)$$

✓ Sympy. Time used: 41.881 (sec). Leaf size: 49

from sympy import * 
t = symbols("t") 
x = Function("x") 
g = Function("g") 
h = Function("h") 
ode = Eq(t*Derivative(x(t), t) + g(t)*x(t) - h(t),0) 
ics = {} 
dsolve(ode,func=x(t),ics=ics)
 

$$(e^{\int \frac{g\left(t\right)}{t}dt}-\int \frac{g\left(t\right)e^{\int \frac{g\left(t\right)}{t}dt}}{t}dt)x\left(t\right)+\int \frac{(g\left(t\right)x\left(t\right)-h\left(t\right))e^{\int \frac{g\left(t\right)}{t}dt}}{t}dt=C_1$$