# Thread: Show that the function y = f(x) is a solution of the given differential equation

1. ## Show that the function y = f(x) is a solution of the given differential equation

$\displaystyle y = \frac{1}{x} \int_1^x\frac {e^t}{t} dt$ , $\displaystyle x^2y' + xy = e^x$

I tried to take the derivative of y equation to get the right equation but I'm keep getting $\displaystyle (x^2)y' = \frac{-e^x}{x}$
Can someone explain this why I can't get to $\displaystyle (x^2)y' + xy = e^x$

2. Originally Posted by fastcarslaugh
$\displaystyle y = \frac{1}{x} \int_1^x\frac {e^t}{t} dt$ , $\displaystyle x^2y' + xy = e^x$

I tried to take the derivative of y equation to get the right equation but I'm keep getting $\displaystyle (x^2)y' = \frac{-e^x}{x}$
Can someone explain this why I can't get to $\displaystyle (x^2)y' + xy = e^x$
You need to use the product rule when you take the derivative of

$\displaystyle \displaystyle y = \underbrace{\frac{1}{x}}_{f(x)} \underbrace{\int_1^x\frac {e^t}{t} dt}_{g(x)}$

So $\displaystyle y'=f'(x)g(x)+f(x)g'(x)$

Now lets compute the parts

$\displaystyle \displaystyle f'(x)=-\frac{1}{x^2}$ and

$\displaystyle \displaystyle g'(x)=\frac{e^x}{x}$ This is from the Fundamental theorem of calculus.

Can you finish from here?

3. But how should I solve the $\displaystyle \int_1^x\frac{e^t}{t}dt$

I am stuck at $\displaystyle y' = \frac{-1}{x^2}\int_1^x\frac{e^t}{t}dt + \frac{e^x}{x^2}$

4. You don't need to evaluate that integral. It will "cancel" when you substitute into the DE.

5. I tried this problem from another perspective.

$\displaystyle y = \frac{1}{x}\int_1^x\frac{e^t}{t}dt$

$\displaystyle xy = \int_1^x\frac{e^t}{t}dt$

$\displaystyle xy' + y = \frac{e^x}{x}$

$\displaystyle x^2y' + xy = e^x$

This method that I used is a lot easier.

Ps. should i make this solved?
Also regarding TheEmptySet post, after looking at it third time that I realized that it just cancels out. So thank you TheEmptySet for your help.

6. Originally Posted by fastcarslaugh
I tried this problem from another perspective.

$\displaystyle y = \frac{1}{x}\int_1^x\frac{e^t}{t}dt$

$\displaystyle xy = \int_1^x\frac{e^t}{t}dt$

$\displaystyle xy' + y = \frac{e^x}{x}$

$\displaystyle x^2y' + xy = e^x$

This method that I used is a lot easier.

Ps. should i make this solved?
Also regarding TheEmptySet post, after looking at it third time that I realized that it just cancels out. So thank you TheEmptySet for your help.
Very elegant!