Need Help?

Subscribe to Differential Equation

  • Notes
  • Comments & Questions

Consider the equation $$y^{\prime}-\frac{1}{3} y=\mathrm{e}^{-t}$$
Find the solution with $$y(0)=a$$
Find the values of a for which (i) $$y \rightarrow 0$$ as $$t \rightarrow \infty$$ (ii) $$y \rightarrow-\infty$$ as $$t \rightarrow \infty$$

(a) $$\frac{d y}{d t}-\frac{1}{3} y=e^{-t} \quad$$ Linear in $$y$$

$$\mu=e^{\int{\frac{-1}{3} d t}}=e^{-\frac{1}{3} t}$$

The general solution: $$\mu y=\int \mu Q(t) d t+c$$

$$e^{-\frac{1}{3} t} \cdot y=\int e^{-\frac{1}{3} t} \cdot e^{-t} d t+c$$

$$\Rightarrow e^{-\frac{1}{3} t} \cdot y=\int e^{-\frac{4}{3} t} d t+c$$

$$e^{-\frac{1}{3} t} \cdot y=-\frac{3}{4} \cdot e^{-\frac{4}{3} t}+c$$

$$e^{\frac{1}{3} t}\left(e^{-\frac{1}{3} t} \cdot y\right)=\left(-\frac{3}{4} \cdot e^{-\frac{4}{3} t}+c\right) e^{\frac{1}{3} t} \Rightarrow$$

$$y=\frac{-3}{4} \cdot e^{-t}+c\: e^{\frac{1}{3} t}$$

$$y(0)=a \Rightarrow$$

$$\frac{-3}{4} e^{0}+c\: e^{0}=-\frac{3}{4}+c=a$$


$$\Rightarrow c=a+\frac{3}{4}$$

$$y=\frac{-3}{4} e^{-t}+\left(a+\frac{3}{4}\right) e^{\frac{1}{3} t}$$

$$\lim _{t \rightarrow \infty} e^{t}=e^{\infty}=\infty$$

$$\lim _{t \rightarrow \infty} e^{-t}=e^{-\infty}=0$$

$$y=-\frac{3}{4} e^{-t}+\left(a+\frac{3}{4}\right) e^{1 / 3 t}$$

(1) $$\lim _{t \rightarrow \infty} y=0 \Rightarrow$$

$$-\frac{3}{4} e^{-\infty}+\left(a+\frac{3}{4}\right) e^{1 / 3(\infty)}=0$$

$$\Rightarrow a=-\frac{3}{4}$$

(2) $$\lim _{t \rightarrow \infty} y=-\infty \Rightarrow$$

$$-\frac{3}{4} e^{-\infty}+\left(a+\frac{3}{4}\right) e^{1 / 3(\infty)}=-\infty$$

$$a+\frac{3}{4}<0 \Rightarrow a<-\frac{3}{4}$$

Find a solution for the equation $$d x-(1+2 x \tan y) d y=0$$

(1) $$d x-(1+2 x \tan y) d y=0 \quad \div d y$$

$$\frac{d x}{d y}-(1+2 x \tan y)=0$$

$$\frac{d x}{d y}-1-2 x \tan y=0$$

$$\frac{d x}{d y}-(2 \tan y) x=+1$$ linear in $$x$$

$$\mu=e^{\int{-2} \tan y d y}=e^{-2 \ln |\sec y|}=\sec ^{-2} y$$

$$\int \tan (y)=\ln |\sec y|$$

$$a \ln |b|=\ln \left|b^{a}\right|$$

$$e^{\ln (x)}=x$$

$$\frac{1}{\sec y}=\cos y$$

$$=\frac{1}{\sec ^{2} y}=\cos ^{2}{y}$$

The general solution:

$$\mu x=\int \mu Q(y) d y+c$$

$$\frac{d x}{d y}-(2 \tan y) x=1$$

$$\cos ^{2} y \cdot x=\int \cos ^{2} y \cdot(1) d y+c$$

$$x \cos ^{2} y=\int\left(\frac{1}{2}+\frac{1}{2} \cos (2 y)\right) d y+c$$

$$\cos ^{2} y=\frac{1}{2}+\frac{1}{2} \cos 2 y$$

$$x \cos ^{2} y=\int \frac{1}{2}(1+\cos 2 y) d y+c \Rightarrow$$

$$x \cos ^{2} y=\frac{1}{2} \cdot\left(y+\frac{1}{2} \sin (2 y)\right)+c$$

No comments yet

Join the conversation

Join Notatee Today!