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Question Number 65858 by ugwu Kingsley last updated on 05/Aug/19

Commented by mathmax by abdo last updated on 05/Aug/19

$$x+2y^{{}^{\prime }}+y={(x+2)}^3\Rightarrow 2y^{{}^{\prime }}+y={(x+2)}^3-x\Rightarrow$$$$2y^{{}^{\prime }}+y=x^3+3x^2\left(2\right)+3x{\left(2\right)}^2+2^3-x\Rightarrow 2y^{{}^{\prime }}+y=x^3+6x^2+11x+8\left(e\right)$$$$\left(he\right)\to 2y^{{}^{\prime }}+y=0\Rightarrow 2y^{{}^{\prime }}=-y\Rightarrow \frac{y^{{}^{\prime }}}{y}=-\frac{1}{2}\Rightarrow ln\mid y\mid =-\frac{x}{2}+\lambda \Rightarrow$$$$y={Ke}^{-\frac{x}{2}}letusemvcmethodwehave{y}^{{}^{\prime }}=K^{{}^{\prime }}{e}^{-\frac{x}{2}}-\frac{K}{2}{e}^{-\frac{x}{2}}$$$$\left(e\right)\Rightarrow 2K^{{}^{\prime }}{e}^{-\frac{x}{2}}-{Ke}^{-\frac{x}{2}}+{Ke}^{-\frac{x}{2}}=x^3+6x^2+11x+8\Rightarrow$$$$K^{{}^{\prime }}=\frac{1}{2}(x^3+6x^2+11x+8)e^{\frac{x}{2}}\Rightarrow K\left(x\right)=\frac{1}{2}\int (x^3+6x^2+11x+8)e^{\frac{x}{2}}dx+c$$$$=\frac{1}{2}\int x^3{e}^{\frac{x}{2}}dx+3\int x^2{e}^{\frac{x}{2}}dx+\frac{11}{2}\int {xe}^{\frac{x}{2}}dx+4\int e^{\frac{x}{2}}dx+c$$$$byparts\int x^3{e}^{\frac{x}{2}}dx=2x^3{e}^{\frac{x}{2}}-\int 3x^2\left(2e^{\frac{x}{2}}\right)dx$$$$=2x^3{e}^{\frac{x}{2}}-6\int x^2{e}^{\frac{x}{2}}dxand\int x^2{e}^{\frac{x}{2}}dx=x^2\left(2e^{\frac{x}{2}}\right)-\int 2x\left(e^{\frac{x}{2}}\right)dx$$$$=2x^2{e}^{\frac{x}{2}}-2\int {xe}^{\frac{x}{2}}dx$$$$\int x{e}^{\frac{x}{2}}dx=x\left(2e^{\frac{x}{2}}\right)-\int \left(2e^{\frac{x}{2}}\right)dx=2x{e}^{\frac{x}{2}}-2\int e^{\frac{x}{2}}dx$$$$=2{xe}^{\frac{x}{2}}-2\left(2e^{\frac{x}{2}}\right)=2{xe}^{\frac{x}{2}}-4e^{\frac{x}{2}}\Rightarrow$$$$\int x^3{e}^{\frac{x}{2}}dx=2x^3{e}^{\frac{x}{2}}-6(2x^2{e}^{\frac{x}{2}}-2(2{xe}^{\frac{x}{2}}-4e^{\frac{x}{2}}))$$$$=\{2x^3-12x^2+24x-48\}{e}^{\frac{x}{2}}\Rightarrow$$$$K\left(x\right)=\{x^3-6x^2+12x-24\}{e}^{\frac{x}{2}}[+3\{2x^2{e}^{\frac{x}{2}}-2(2{xe}^{\frac{x}{2}}-4e^{\frac{x}{2}})\}$$$$+\frac{11}{2}\{2{xe}^{\frac{x}{2}}-4e^{\frac{x}{2}}\}+4\left\{2e^{\frac{x}{2}}\right\}+c$$$$y=K\left(x\right)e^{-\frac{x}{2}}\Rightarrow y\left(x\right)=...$$

Commented by mathmax by abdo last updated on 05/Aug/19

$$2)x^2{y}^{{}^{\prime }}+3xy=x^3+2x^2\Rightarrow {xy}^{{}^{\prime }}+3y=x^2+2x\left(e\right)$$$$\left(he\right)\Rightarrow {xy}^{{}^{\prime }}+3y=0\Rightarrow {xy}^{{}^{\prime }}=-3y\Rightarrow \frac{y^{{}^{\prime }}}{y}=\frac{-3}{x}\Rightarrow ln\mid y\mid =-3ln\mid x\mid +\lambda \Rightarrow$$$$y\left(x\right)=\frac{k}{\mid x{\mid }^3}letfindthesolutionon]0,+\mathrm{\infty }[\Rightarrow y\left(x\right)=\frac{k}{x^3}$$$$mvcmethod\to y^{{}^{\prime }}=\frac{k^{{}^{\prime }}}{x^3}+k(-3\frac{x^2}{x^6})=\frac{k^{{}^{\prime }}}{x^3}-\frac{3k}{x^4}$$$$$$$$\left(e\right)\Rightarrow \frac{k^{{}^{\prime }}}{x^2}-\frac{3k}{x^3}+\frac{3k}{x^3}=x^2+2x\Rightarrow k^{{}^{\prime }}=x^4+3x^3\Rightarrow k\left(x\right)=\int (x^4+3x^3)dx+c$$$$\Rightarrow k\left(x\right)=\frac{x^5}{5}+\frac{3}{4}{x}^4+c\Rightarrow y\left(x\right)=\frac{1}{x^3}\{\frac{x^5}{5}+\frac{3}{4}{x}^4+c\}$$$$=\frac{1}{5}{x}^2+\frac{3}{4}x+\frac{c}{x^3}.$$

Commented by mathmax by abdo last updated on 05/Aug/19

$${xy}^{{}^{\prime }}-y=cos\left(4x\right)$$$$\left(he\right)\Rightarrow {xy}^{{}^{\prime }}-y=0\Rightarrow {xy}^{{}^{\prime }}=y\Rightarrow \frac{y^{{}^{\prime }}}{y}=\frac{1}{x}\Rightarrow ln\mid y\mid =ln\mid x\mid +\lambda \Rightarrow$$$$y=k\mid x\mid letfindthesolutionon]0,+\mathrm{\infty }[\Rightarrow y=kx$$$$\Rightarrow y^{{}^{\prime }}=k^{{}^{\prime }}x+k$$$$\left(e\right)\Rightarrow k^{{}^{\prime }}{x}^2+kx-kx=cos\left(4x\right)\Rightarrow k^{{}^{\prime }}{x}^2=cos\left(4x\right)\Rightarrow$$$$k^{{}^{\prime }}=\frac{cos\left(4x\right)}{x^2}\Rightarrow k\left(x\right)=\int \frac{cos\left(4x\right)}{x^2}dx+cbyparts$$$$\int \frac{cos\left(4x\right)}{x^2}dx=-\frac{1}{x}cos\left(4x\right)-\int \left(\frac{1}{x}\right)4sin\left(4x\right)dx$$$$=-\frac{cos\left(4x\right)}{x}-4\int \frac{sin\left(4x\right)}{x}dx\Rightarrow k\left(x\right)=-\frac{cos\left(4x\right)}{x}-4\int \frac{sin\left(4x\right)}{x}dx+c$$$$\Rightarrow y\left(x\right)=-cos\left(4x\right)-4x{\int }^x\frac{sin\left(4t\right)}{t}dt+cx$$

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