Question

Giải phương trình y"= lnx

Answer 1

Integrate both sides with respect to x :

`y''(x) = \ln(x) \implies \displaystyle \int y''(x)\,\mathrm dx = \int \ln(x)\,\mathrm dx`

On the right side, integrate by parts with

`f = \ln(x) \implies \mathrm df = \frac{\mathrm dx}x \\\\ \mathrm dg = \mathrm dx \implies g = x`

`\implies \displaystyle \int\ln(x)\,\mathrm dx = fg - \int g\,\mathrm df \\\\ \int\ln(x)\,\mathrm dx= x\ln(x) - \int \mathrm dx \\\\ \int\ln(x)\,\mathrm dx= x\ln(x) - x + C_1`

Then

`\displaystyle y'(x) = x\ln(x) - x + C_1`

Integrate both sides with respect to x by parts again, this time with

`f = \ln(x) \implies \mathrm df = \frac{\mathrm dx}x \\\\ \mathrm dg = x\,\mathrm dx \implies g = \frac{x^2}2`

`\implies \displaystyle \int x\ln(x)\,\mathrm dx = fg-\int g\,\mathrm df \\\\ \int x\ln(x)\,\mathrm dx = \frac{x^2\ln(x)}2 - \int\frac x2\,\mathrm dx \\\\ \int x\ln(x)\,\mathrm dx = \frac{x^2\ln(x)}2-\frac{x^2}4 + C_2`

Then

`y(x) = \frac{x^2\ln(x)}2 - \frac{x^2}4 - \frac{x^2}2 + C_1x + C_2 \\\\ \boxed{y(x) = \frac{(2\ln(x)-3)x^2}4 + C_1x + C_2}`