Question #288877

Solve :


x²y" -2xy' -4y=x²+2ln x

1
Expert's answer
2022-01-30T16:47:42-0500

Let us solve the differential equation x2y2xy4y=x2+2lnx.x^2y'' -2xy' -4y=x^2+2\ln x.

Let us use the transformation x=et.x=e^t. Then yx=ytet, yx2=(yt2yt)e2t.y'_x=y'_te^{-t},\ y''_{x^2}=(y''_{t^2}-y'_t)e^{-2t}.

We get the equation e2t(yt2yt)e2t2etytet4y=e2t+2t,e^{2t}(y''_{t^2}-y'_t)e^{-2t} -2e^ty'_te^{-t} -4y=e^{2t}+2t, which is equivalent to yt23yt4y=e2t+2t.y''_{t^2}-3y'_t -4y=e^{2t}+2t.

The characteristic equation k23k4=0k^2-3k-4=0 of the differential equation yt23yt4y=0y''_{t^2}-3y'_t -4y=0 is equivalent to (k+1)(k4)=0,(k+1)(k-4)=0, and hence has the roots k1=1, k2=4.k_1=-1,\ k_2=4.

It follows that the general solution of the differential equation yt23yt4y=e2t+2ty''_{t^2}-3y'_t -4y=e^{2t}+2t is y(t)=C1et+C2e4t+yp,y(t)=C_1e^{-t}+C_2e^{4t}+y_p, where yp=ae2t+bt+d.y_p=ae^{2t}+bt+d. Then yp=2ae2t+b, yp=4ae2t.y_p'=2ae^{2t}+b,\ y_p''=4ae^{2t}.

It follows that

4ae2t3(2ae2t+b)4(ae2t+bt+d)=e2t+2t,4ae^{2t}-3(2ae^{2t}+b)-4(ae^{2t}+bt+d)=e^{2t}+2t, and hence

6ae2t4bt3b4d=e2t+2t.-6ae^{2t}-4bt-3b-4d=e^{2t}+2t.

We conclude that 6a=1, 4b=2, 3b4d=0,-6a=1,\ -4b=2,\ -3b-4d=0, and thus a=16, b=12, d=34b=38.a=-\frac{1}6,\ b=-\frac{1}2,\ d=-\frac{3}4b=\frac{3}8.

Therefore, the general solution of the differential equation yt23yt4y=e2t+2ty''_{t^2}-3y'_t -4y=e^{2t}+2t is y(t)=C1et+C2e4t16e2t12t+38.y(t)=C_1e^{-t}+C_2e^{4t}-\frac{1}6e^{2t}-\frac{1}2t+\frac{3}8.

We conclude that the general solution of the differential equation x2y2xy4y=x2+2lnxx^2y'' -2xy' -4y=x^2+2\ln x is y(x)=C1x1+C2x416x212lnx+38.y(x)=C_1x^{-1}+C_2x^{4}-\frac{1}6x^{2}-\frac{1}2\ln x+\frac{3}8.



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