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

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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