Question

verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of denition of each solution. 25. x dy dx 2 3xy 5 1; y 5 e3x # x 1 e23t t d

Answer 1

$$x\dfrac{dy}{dx}-3xy=1$$

$$y=e^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt$$

$$\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(e^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt\bigg)$$

$$=3e^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt+e^{3x}\dfrac{e^{-3x}}{x}$$

$$=3e^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt+\dfrac{1}{x}$$

Substitute

$$x\dfrac{dy}{dx}-3xy=1$$

$$x\bigg(3e^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt+\dfrac{1}{x}\bigg)$$

$$-3x\bigg(e^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt\bigg)=1$$

$$3xe^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt+1-3xe^{3x}\displaystyle\int_{1}^{x}\dfrac{e^{-3t}}{t}dt=1$$

$$1=1, True$$

Therefore the indicated function is a solution of the given differential equation.