`dy/(dt) = t^2/sqrt(3+5t)` Solve the differential equation.

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`dy/dt=t^2/sqrt(3+5t)`

`y=intt^2/sqrt(3+5t)dt`

Apply integral substitution :`u=sqrt(3+5t)`

`du=1/2(3+5t)^(1/2-1)(5)dt`

`du=5/(2sqrt(3+5t))dt`

`=>dt/sqrt(3+5t)=2/5du`

`u=sqrt(3+5t)`

squaring above,

`u^2=3+5t`

`=>5t=u^2-3`

`t=(u^2-3)/5`

`t^2=1/25(u^2-3)^2`

`t^2=1/25(u^4-6u^2+9)`

`y=intt^2/(sqrt(3+5t))dt`

`=int1/25(u^4-6u^2+9)(2/5)du`

`=int2/125(u^4-6u^2+9)du`

Take the constant out,

`=2/125int(u^4-6u^2+9)du`

Apply the sum and power rule,

`=2/125(intu^4du-int6u^2du+int9du)`

`=2/125(u^5/5-6u^3/3+9u)`

`=2/125(u^5/5-2u^3+9u)`

Substitute back `u=sqrt(3+5t)`

`=2/125(1/5(3+5t)^(5/2)-2(3+5t)^(3/2)+9(3+5t)^(1/2))`

simplify the above and add a constant C to the solution,

`=2/125(3+5t)^(1/2)(1/5(3+5t)^2-2(3+5t)+9)+C`

`=2/125sqrt(3+5t)(1/5(9+25t^2+30t)-2(3+5t)+9)+C`

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`dy/dt=t^2/sqrt(3+5t)`

`y=intt^2/sqrt(3+5t)dt`

Apply integral substitution :`u=sqrt(3+5t)`

`du=1/2(3+5t)^(1/2-1)(5)dt`

`du=5/(2sqrt(3+5t))dt`

`=>dt/sqrt(3+5t)=2/5du`

`u=sqrt(3+5t)`

squaring above,

`u^2=3+5t`

`=>5t=u^2-3`

`t=(u^2-3)/5`

`t^2=1/25(u^2-3)^2`

`t^2=1/25(u^4-6u^2+9)`

`y=intt^2/(sqrt(3+5t))dt`

`=int1/25(u^4-6u^2+9)(2/5)du`

`=int2/125(u^4-6u^2+9)du`

Take the constant out,

`=2/125int(u^4-6u^2+9)du`

Apply the sum and power rule,

`=2/125(intu^4du-int6u^2du+int9du)`

`=2/125(u^5/5-6u^3/3+9u)`

`=2/125(u^5/5-2u^3+9u)`

Substitute back `u=sqrt(3+5t)`

`=2/125(1/5(3+5t)^(5/2)-2(3+5t)^(3/2)+9(3+5t)^(1/2))`

simplify the above and add a constant C to the solution,

`=2/125(3+5t)^(1/2)(1/5(3+5t)^2-2(3+5t)+9)+C`

`=2/125sqrt(3+5t)(1/5(9+25t^2+30t)-2(3+5t)+9)+C`

`=2/125sqrt(3+5t)((9+25t^2+30t-10(3+5t)+45)/5)+C`

`=2/125sqrt(3+5t)((9+25t^2+30t-30-50t+45)/5)+C`

`=2/125sqrt(3+5t)((25t^2-20t+24)/5)+C`

`y=2/625(25t^2-20t+24)sqrt(3+5t)+C`

 

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