Answer to Question #117715 in Differential Equations for samuel abdu

Question

Solve the initial value problem $2f''(t)-3f'(t)+f(t)=te^t, f(0)=1,f'(0)=\frac{3}{2}$

Using the point transformation [y = exp(ψ)], transform the equation (Eହ

) into a first

order non-linear O.D.E with independent variable x and dependent variable w, whereby w ≔ ψ

ᇱ

.

Afterwards, use the second point transformation ቂw =

ିଶఉᇲ(௫)

ఉ

ቃ to change this first order O.D.E to a

second order, linear and homogeneous O.D.E with independent variable x and dependent variable β .

Answer 1

The solution consists of two parts:

$$f(t)=f_{hom}(t)+f_{par}(t)\\[0.3cm] f_{hom}(t)-\text{solution of a homogeneous equation}\\[0.3cm] 2f''(t)-3f'(t)+f(t)=0\\ f_{par}(t)-\text{particular solution of an inhomogeneous equation}\\[0.3cm] 2f''(t)-3f'(t)+f(t)=t\cdot e^t$$

1 STEP: We solve the homogeneous equation

$$2f''(t)-3f'(t)+f(t)=0$$

We will seek a solution in the form $f(t)=e^{kt}$ , then

$$f(t)=e^{kt}\longrightarrow \left\{\begin{array}{l} f'(t)=k\cdot e^{kt}\\[0.3cm] f''(t)=k^2\cdot e^{kt} \end{array}\right.\longrightarrow\\[0.3cm] 2\cdot k^2\cdot e^{kt}-3\cdot k\cdot e^{kt}+e^{kt}=0\\[0.3cm] e^{kt}\cdot\left(2k^2-3k+1\right)=0\longrightarrow\underbrace{2k^2-3k+1=0}_{\text{quadratic equation}}\\[0.3cm] \sqrt{D}=\sqrt{\left(-3\right)^2-4\cdot 2\cdot1}=\sqrt{1}=1\\[0.3cm] \left[\begin{array}{l} k_1=\displaystyle\frac{-(-3)+1}{2\cdot2}=1\\[0.3cm] k_2=\displaystyle\frac{-(-3)-1}{2\cdot2}=\displaystyle\frac{1}{2} \end{array}\right.\\[0.3cm] \boxed{f_{hom}(t)=C_1\cdot e^t+C_2\cdot e^{t/2}}$$

2 STEP: We solve the inhomogeneous equation

$$2f''(t)-3f'(t)+f(t)=t\cdot e^t$$

Minimum required theory: when solving inhomogeneous equations of the form

$$F\left(f,f',f'',f''',\ldots\right)=g(t)$$

if the right side is $g(t)=P_m(t)\cdot e^{\alpha t},\,\,\text{where}\,\,\, P_m(t)-\text{polynomial}$ , then to look for the particular solution in the form

$$\left[\begin{array}{l} \widetilde{P_m}(t)\cdot e^{\alpha t}-\alpha\,\,\text{not the root of the characteristic equation}\\[0.3cm] x^s\widetilde{P_m}(t)\cdot e^{\alpha t}-\alpha\,\,\text{the root of the characteristic equation} \end{array}\right.$$

In our case,

$$g(t)=t\cdot e^{1\cdot t}\longrightarrow k_1=1-\text{multiplicity root 1}\\[0.3cm] f_{par}(t)=t\cdot\left(At+B\right)\cdot e^t\equiv\left(At^2+Bt\right)\cdot e^t\longrightarrow\\[0.3cm] \left\{\begin{array}{l} f'(t)=\left(At^2+Bt\right)\cdot e^t+\left(2At+B\right)\cdot e^t\\[0.3cm] f'(t)=\left(At^2+(2A+B)t+B\right)\cdot e^t\\[0.3cm] f''(t)=\left(At^2+(2A+B)t+B\right)\cdot e^t+\left(2At+(2A+B)\right)\cdot e^t\\[0.3cm] f''(t)=\left(At^2+(4A+B)t+(2A+2B)\right)\cdot e^t \end{array}\right.$$

Substitute in the original equation

$$2\left(At^2+(4A+B)t+(2A+2B)\right)\cdot e^t-\\[0.3cm] -3\left(At^2+(2A+B)t+B\right)\cdot e^t+\left(At^2+Bt\right)\cdot e^t=t\cdot e^t\\[0.3cm] (2A-3A+A)t^2+(2(4A+B)-3(2A+B)+B)t+\\[0.3cm] +(2(2A+2B)-3B)=t\\[0.3cm] 0\cdot t^2+(2A)\cdot t+(4A+B)=t\longrightarrow\\[0.3cm] \left\{\begin{array}{l} 2A=1\\[0.3cm] 4A+B=0 \end{array}\right. \longrightarrow \left\{\begin{array}{l} A=\displaystyle\frac{1}{2}\\[0.3cm] B=-4A=-2 \end{array}\right.$$

Conclusion,

$$\boxed{f_{par}(t)=\left(\frac{t^2}{2}-2t\right)\cdot e^t}$$

General solution,

$$\boxed{f(t)=C_1\cdot e^t+C_2\cdot e^{t/2}+\left(\frac{t^2}{2}-2t\right)\cdot e^t}$$

The constants are determined from the initial conditions

$$f(0)=1=C_1\cdot e^0+C_2\cdot e^{0/2}+\left(\frac{0^2}{2}-2\cdot0\right)\cdot e^0\\[0.3cm] \boxed{C_1+C_2=1}\\[0.3cm] f'(t)=C_1\cdot e^t+\frac{C_2}{2}\cdot e^{t/2}+(2t-2)\cdot e^t+\left(\frac{t^2}{2}-2t\right)\cdot e^t\\[0.3cm] f'(0)=\frac{3}{2}=C_1\cdot e^0+\frac{C_2}{2}\cdot e^{0/2}+(2\cdot0-2)\cdot e^0+\left(\frac{0^2}{2}-2\cdot0\right)\cdot e^0\\[0.3cm] \boxed{C_1+\frac{C_2}{2}-2=\frac{3}{2}}$$

It remains to solve the system of equations

$$\left\{\begin{array}{l} C_1+C_2=1\\[0.3cm] \left.C_1+\displaystyle\frac{C_2}{2}-2=\displaystyle\frac{3}{2}\right|\times(2) \end{array}\right.\longrightarrow\left\{\begin{array}{l} C_2=1-C_1\\[0.3cm] 2C_1+C_2-4=3\end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} C_2=1-C_1\\[0.3cm] 2C_1+1-C_1-4=3\end{array}\right.\longrightarrow \left\{\begin{array}{l} C_2=-5\\[0.3cm] C_1=6\end{array}\right.$$

Conclusion, solution of IVT is

$$\boxed{f(t)=6\cdot e^t-5\cdot e^{t/2}+\left(\frac{t^2}{2}-2t\right)\cdot e^t}$$

ANSWER

$$f(t)=6\cdot e^t-5\cdot e^{t/2}+\left(\frac{t^2}{2}-2t\right)\cdot e^t$$