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Giải phương trình y"= lnx

User Arikael
by
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1 Answer

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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}

User Divakar
by
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