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Question Number 87532 by niroj last updated on 04/Apr/20

$$\left(1\right).\mathbf{Find}\mathbf{the}\mathbf{general}\mathbf{solution}:$$$$\mathbf{y}=\mathbf{px}+{\mathbf{p}}^{\mathbf{n}}$$$$\left(2\right).\mathbf{Solve}\mathbf{the}\mathbf{differential}\mathbf{equation}:$$$${(\mathbf{x}+1)}^2\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}+(\mathbf{x}+1)\frac{\mathbf{dy}}{\mathbf{dx}}=(2\mathbf{x}+3)(2\mathbf{x}+4).$$$$$$

Commented by mathmax by abdo last updated on 07/Apr/20

$$2){(x+1)}^2{y}^{{}^{″}}+(x+1)y^{{}^{\prime }}=(2x+3)(2x+4)\Rightarrow$$$$(x+1)y^{{}^{″}}+y^{{}^{\prime }}=\frac{4x^2+8x+6x+12}{x+1}=\frac{4x^2+14x+12}{x+1}$$$$let{y}^{{}^{\prime }}=z\left(e\right)\Rightarrow (x+1)z^{{}^{\prime }}+z=\frac{4x^2+14x+12}{x+1}$$$$\left(he\right)\to (x+1)z^{{}^{\prime }}+z=0\Rightarrow (x+1)z^{{}^{\prime }}=-z\Rightarrow \frac{z^{{}^{\prime }}}{z}=-\frac{1}{x+1}\Rightarrow$$$$ln\mid z\mid =-ln\mid x+1\mid +c\Rightarrow z=\frac{k}{\mid x+1\mid }letdeterminethesolutionon$$$$]-1,+\mathrm{\infty }[\Rightarrow z\left(x\right)=\frac{k}{x+1}mvcmethod\to z^{{}^{\prime }}=-\frac{1}{{(x+1)}^2}k+\frac{1}{x+1}{k}^{{}^{\prime }}$$$$\left(e\right)\Rightarrow -\frac{1}{x+1}k+k^{{}^{\prime }}+\frac{k}{x+1}=\frac{4x^2+14x+12}{x+1}\Rightarrow$$$$k\left(x\right)=\int \frac{4x^2+14x+12}{x+1}dx+c$$$$=\int \frac{4(x^2-1)+14x+16}{x+1}dx+c=\int 4(x-1)+\int \frac{14x+16}{x+1}dx+c$$$$=2{(x-1)}^2+\int \frac{14(x+1)-14+16}{x+1}dx+c$$$$=2{(x-1)}^2+14x+2ln(x+1)+c\Rightarrow$$$$z\left(x\right)=\frac{k\left(x\right)}{x+1}=\frac{1}{x+1}(2x^2-4x+2+14x+2ln(x+1)+c)$$$$=\frac{1}{x+1}(2x^2+10x+2ln(x+1)+c+2)$$$$wehave{y}^{{}^{\prime }}=z\Rightarrow y\left(x\right)=\int \frac{2x^2+10x+2ln(x+1)+\lambda }{x+1}dx$$$$(\lambda =c+2)$$

Commented by niroj last updated on 03/May/20

thanks for effort��

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