verify that each given function is a solution of the differential equation ty′−y=t2;y=3t+t2

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lamusesamuset · Accepted Answer

Step 1We have differentiated y(t)y′(t)=dydty′(t)=3+2tStep 2Plug the values of y(t) and y′(t) in the given differential equationt(3+2t)−(3t+t2)−t23t+2t2−3t−t2−t2y1(t) has satisfied the equation Thus y1(t) is a solution of given differential equation.Resulty is a solution of given differential equation

RizerMix · Accepted Answer

Step 1:First, let&#039;s find the derivative of y with respect to t. Using the power rule, we have:dydt=ddt(3t+t2)=3+2tStep 2:Next, we substitute y and dydt into the differential equation:t(dydt)−y=t(3+2t)−(3t+t2)=3t+2t2−3t−t2=t2We observe that the expression simplifies to t2, which is indeed equal to the right-hand side of the differential equation.Since the left-hand side of the differential equation, ty′−y, equals the right-hand side, t2, we can conclude that the function y=3t+t2 is indeed a solution of the differential equation ty′−y=t2.

user_27qwe · Accepted Answer

Let&#039;s start by substituting y=3t+t2 into the differential equation:t·dydt−y=t2Now, let&#039;s calculate the derivative of y with respect to t. The derivative of 3t is 3, and the derivative of t2 is 2t. Applying the power rule of differentiation, we have:dydt=ddt(3t+t2)=3+2tSubstituting this derivative back into the equation, we get:t·(3+2t)−(3t+t2)=t2Expanding the terms, we have:3t+2t2−3t−t2=t2Simplifying the equation, we find:t2=t2Since the left side of the equation is equal to the right side, we can conclude that y=3t+t2 is indeed a solution to the given differential equation ty′−y=t2.Therefore, the function y=3t+t2 satisfies the differential equation.

karton · Accepted Answer

Answer:y=3t+t2Explanation:To verify that each given function is a solution of the differential equation ty′−y=t2, we need to substitute the given function y into the equation and see if it satisfies the equation for all values of t.Let&#039;s start by substituting y=3t+t2 into the differential equation:t(dydt)−(3t+t2)=t2Now, we need to differentiate y with respect to t to find dydt:y=3t+t2⇒dydt=3+2tSubstituting this back into the equation:t·(3+2t)−(3t+t2)=t2Simplifying this equation gives us:3t+2t2−3t−t2=t2⇒t2=t2As we can see, both sides of the equation are equal, so the given function y=3t+t2 is indeed a solution of the differential equation ty′−y=t2.Therefore, we have verified that y=3t+t2 is a solution of the differential equation.

verify that each given function is a solution of the differential equation ty

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