Find the particular anti-derivative of dy/dx = 3/x + 1/x^2, y(1) =1

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violy's profile pic

Posted on

We will use integration here.

`intdy = int(3/x + 1/x^2)dx `

`intdy = int(3/x + 1x^-2)dx`

`y = 3lnx -1/x + c` 

Using the formulas: `int(du/u) = lnu + c` and `intx^ndx = x^(n+1)/(n+1) + c`

Now, we will use the given y(1) = 1.

`1 = 3ln(1) - 1/(1) + c`

`1 = 3(0) - 1 + c`

`1 = 0 - 1 + c`

Add 1 on both sides. 

`2 = c`

So, the final answer will be: 

y = 3lnx -1/x + 2 or y = (3xlnx +2x - 1)/x

oldnick's profile pic

Posted on

`dy/dx=3/x+1/x^2 `        `y(1)=1`

 

`y=3logx-1/x +c`

for: `d/dx( 3logx-1/x+c)=3/x-(-1)x^(-2)=3/x+1/x^2`

since `y(1)=1 ` : then:   `3log(1) -1+c=1`  `rArr c=2`

`y= 3logx -1/x+2`   (particular integral)

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