Question #300202

Use appropriate substitution to reduce the following equation to a variable separable and then solve the given IVP: dy/dx=(3x+2y)/(3x+2y+2),y(1)=1dy/dx= (3x+2y)/(3x+2y+2), y(-1)=-1

1
Expert's answer
2022-02-21T16:43:53-0500

Let us solve the differential equation

dydx=3x+2y3x+2y+2,  y(1)=1.\frac{dy}{dx}=\frac{3x+2y}{3x+2y+2},\ \ y(−1)=−1.

Let us use the transformation u=3x+2y.u=3x+2y. Then dudx=3+2dydx,\frac{du}{dx}=3+2\frac{dy}{dx}, and hence dydx=12(dudx3).\frac{dy}{dx}=\frac{1}2(\frac{du}{dx}-3).

It follows that

12(dudx3)=uu+2,\frac{1}2(\frac{du}{dx}-3)=\frac{u}{u+2}, and hence dudx=2uu+2+3=5u+6u+2.\frac{du}{dx}=\frac{2u}{u+2}+3=\frac{5u+6}{u+2}.

We get the equation dx=(u+2)du5u+6,{dx}=\frac{(u+2)du}{5u+6}, and hence dx=(u+2)du5u+6=15(5u+10)du5u+6=15(1+45u+6)du=u5+425ln5u+6+C.\int{dx}=\int\frac{(u+2)du}{5u+6}=\frac{1}5\int\frac{(5u+10)du}{5u+6} =\frac{1}5\int(1+\frac{4}{5u+6})du=\frac{u}5+\frac{4}{25}\ln|5u+6|+C.

Therefore,

x=3x+2y5+425ln5(3x+2y)+6+Cx=\frac{3x+2y}5+\frac{4}{25}\ln|5(3x+2y)+6|+C

Since y(1)=1,y(-1)=-1, we conclude that 1=1+425ln19+C,-1=-1+\frac{4}{25}\ln 19+C, and thus C=425ln19.C=-\frac{4}{25}\ln 19.

We conclude that the solution of dydx=3x+2y3x+2y+2,  y(1)=1\frac{dy}{dx}=\frac{3x+2y}{3x+2y+2},\ \ y(−1)=−1 is

x=3x+2y5+425ln15x+10y+6425ln19.x=\frac{3x+2y}5+\frac{4}{25}\ln|15x+10y+6|-\frac{4}{25}\ln 19.


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