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Question Number 90863 by ajfour last updated on 26/Apr/20

$$\frac{dy}{dx}-a\left(\frac{y}{x}\right)=1+\frac{1}{x}$$

Answered by mr W last updated on 26/Apr/20

$$p\left(x\right)=-\frac{a}{x}$$$$\int p\left(x\right)dx=-\int \frac{a}{x}dx=-a\ln x=\ln x^{-a}$$$$u\left(x\right)=e^{\int p\left(x\right)dx}=x^{-a}$$$$q\left(x\right)=1+\frac{1}{x}$$$$\int u\left(x\right)q\left(x\right)dx=\int x^{-a}(1+\frac{1}{x})dx=-\frac{1}{a-1}{x}^{-a+1}-\frac{1}{a}{x}^{-a}$$$$y=\frac{\int u\left(x\right)q\left(x\right)dx+C}{u\left(x\right)}=\frac{-\frac{1}{a-1}{x}^{-a+1}-\frac{1}{a}{x}^{-a}+C}{x^{-a}}$$$$\Rightarrow y={Cx}^a-\frac{x}{a-1}-\frac{1}{a}$$

Commented by jagoll last updated on 26/Apr/20

$$exactdiffeq?$$

Commented by mr W last updated on 26/Apr/20

Commented by jagoll last updated on 26/Apr/20

$$oyes.notexact$$

Commented by jagoll last updated on 26/Apr/20

Commented by ajfour last updated on 26/Apr/20

Absolutely perfect Sir!

Answered by mathmax by abdo last updated on 23/Jun/20

$${\mathrm{y}}^{{}^{\prime }}-\frac{\mathrm{a}}{\mathrm{x}}\mathrm{y}=\frac{\mathrm{x}+1}{\mathrm{x}}\Rightarrow {\mathrm{xy}}^{{}^{\prime }}-\mathrm{ay}=\mathrm{x}+1\left(\mathrm{e}\right)$$$$\left(\mathrm{he}\right)\to {\mathrm{xy}}^{{}^{\prime }}-\mathrm{ay}=0\Rightarrow {\mathrm{xy}}^{{}^{\prime }}=\mathrm{ay}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{\mathrm{a}}{\mathrm{x}}\Rightarrow \ln \mid \mathrm{y}\mid =\mathrm{aln}\mid \mathrm{x}\mid +\mathrm{c}\Rightarrow$$$$\mathrm{y}=\mathrm{k}\mid \mathrm{x}{\mid }^{\mathrm{a}}\mathrm{solution}\mathrm{on}]0,+\mathrm{\infty }[\Rightarrow \mathrm{y}=\mathrm{k}{\mathrm{x}}^{\mathrm{a}}$$$$\mathrm{mvc}\mathrm{method}\to {\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}{\mathrm{x}}^{\mathrm{a}}+\mathrm{ak}{\mathrm{x}}^{\mathrm{a}-1}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{x}}^{\mathrm{a}+1}+\mathrm{ak}{\mathrm{x}}^{\mathrm{a}}-\mathrm{ka}{\mathrm{x}}^{\mathrm{a}}=\mathrm{x}+1\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\frac{\mathrm{x}+1}{{\mathrm{x}}^{\mathrm{a}+1}}=\frac{1}{{\mathrm{x}}^{\mathrm{a}}}+\frac{1}{{\mathrm{x}}^{\mathrm{a}+1}}={\mathrm{x}}^{-\mathrm{a}}+{\mathrm{x}}^{-(\mathrm{a}+1)}\Rightarrow$$$$\mathrm{k}\left(\mathrm{x}\right)=\int ({\mathrm{x}}^{-\mathrm{a}}+{\mathrm{x}}^{-(\mathrm{a}+1)})\mathrm{dx}=\frac{1}{1-\mathrm{a}}{\mathrm{x}}^{1-\mathrm{a}}-\frac{1}{\mathrm{a}}{\mathrm{x}}^{-\mathrm{a}}+\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=(\frac{1}{1-\mathrm{a}}{\mathrm{x}}^{1-\mathrm{a}}-\frac{1}{\mathrm{a}}{\mathrm{x}}^{-\mathrm{a}}){\mathrm{x}}^{\mathrm{a}}=\frac{\mathrm{x}}{1-\mathrm{a}}-\frac{1}{\mathrm{a}}\mathrm{with}\mathrm{a}\ne 1\mathrm{and}\mathrm{a}\ne 0$$

Commented by mathmax by abdo last updated on 23/Jun/20

$$\mathrm{sorry}\mathrm{y}\left(\mathrm{x}\right)=(\frac{1}{1-\mathrm{a}}{\mathrm{x}}^{1-\mathrm{a}}-\frac{1}{\mathrm{a}}{\mathrm{x}}^{-\mathrm{a}}+\mathrm{c}){\mathrm{x}}^{\mathrm{a}}=\frac{\mathrm{x}}{1-\mathrm{a}}-\frac{1}{\mathrm{a}}+\mathrm{c}{\mathrm{x}}^{\mathrm{a}}$$

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