I have two problems for which I know the answers (and working) but am still confused about the method used to solve them. The equations are yy(2)=(y(1))2 and x(2)+(x(1))2=0In my notes it instructs that in the case of the independent variable missing from the equation (x and t, respectively), the substitution to reduce the order is made as follows:p=y(1)y(2)=dpdx=dpdydydx=pdpdyThen, for the first equation:yy(2)=(y(1))2⇒ypdpdy=p2But for the second equation, the substitution is made as:x(2)+(x(1))2=0⇒dpdt=−p2I've tested this with online calculators and even they compute these two differential equations in the two different ways.Any explanation of where I'm going wrong would be greatly appreciated.

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I have two problems for which I know the answers (and working) but am still confused about the method used to solve them. The equations are yy(2)=(y(1))2 and x(2)+(x(1))2=0In my notes it instructs that in the case of the independent variable missing from the equation (x and t, respectively), the substitution to reduce the order is made as follows:p=y(1)y(2)=dpdx=dpdydydx=pdpdyThen, for the first equation:yy(2)=(y(1))2⇒ypdpdy=p2But for the second equation, the substitution is made as:x(2)+(x(1))2=0⇒dpdt=−p2I&#039;ve tested this with online calculators and even they compute these two differential equations in the two different ways.Any explanation of where I&#039;m going wrong would be greatly appreciated.

sorrisi7yny · Accepted Answer

Step 1Both substitutions are correct and give the same answer but the first substitution is easier to integrate:ddpdt=−p2−dp′p2=1(d1pright)′=1d1p=t+cp=d1t+cx+c2=ln⁡(t+c)c1ex=t+cStep 2For the second substitution:pddpdx=−p2ddpp=−dxln⁡p=−x+cp=Ce−xx′ex=Cex=c1t+c2Step 3This differential equation has no term in x so you can also substitute x″=ddpdt. In the first differential equation you have the factor y. So if you try to substitute y″=ddpdt;yy″=y′2yddpdt=p2You can&#039;t integate this since you have p,t,y so you have to substitute ypdpdy.

I have two problems for which I know

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